Class 12: Three Dimensional Geometry

$\displaystyle \textbf{Question 1: } \text{If a line makes an angle } \alpha,\beta \text{ and } \gamma \text{ with positive direction of the }$
$\displaystyle \text{coordinate axes, then the value of} \sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma \text{ will be } \quad \text{ISC 2024}$
$\displaystyle \text{(a) } 1 \quad \text{(b) } 3 \quad \text{(c) } -2 \quad \text{(d) } 2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d)}\ \text{We have, }\sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma$
$\displaystyle =(1-\cos^{2}\alpha)+(1-\cos^{2}\beta)+(1-\cos^{2}\gamma)$
$\displaystyle =3-(\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma)$
$\displaystyle =3-1=2\ \ [\because \cos\alpha,\cos\beta,\cos\gamma\text{ are direction cosines}]$
$\\$

$\displaystyle \textbf{Question 2: } \text{In the figure given below, if the coordinates of the } \text{point } P \text{ are } (a,b,c),$
$\displaystyle \text{ then what are the perpendicular} \text{distances of } P \text{ from } XY, YZ \text{ and } ZX \text{ planes}$
$\displaystyle \text{respectively?} \quad \text{ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }PD,PE,PF\text{ be perpendicular distances of }P\text{ from }XY,YZ,ZX\text{-planes}$
$\displaystyle \text{For }P(a,b,c),\ a,b,c\text{ are perpendicular distances from }YZ,ZX,XY\text{-planes respectively}$
$\displaystyle \therefore \text{Required distances from }XY,YZ,ZX\text{-planes are }c,a,b\text{ respectively}$
$\\$

$\displaystyle \textbf{Question 3: } \text{A mobile tower is situated at the top of a hill. Consider the surface on }$
$\displaystyle \text{which the tower stands as a plane having } \text{points } A(1,0,2), B(3,-1,1) \text{ and } C(1,2,1)$
$\displaystyle \text{ on it. The mobile } \text{tower is tied with three cables from the points } A,B \text{ and }$
$\displaystyle C \text{ such }\text{that it stands vertically on the ground. The top of the }\text{tower is at point } P(2,3,1)$
$\displaystyle \text{ as shown in the figure below. } \text{The foot of the perpendicular from the point } P \text{ on the}$
$\displaystyle \text{plane is at the point } Q\left(\frac{43}{29},\frac{77}{29},\frac{9}{29}\right). \quad \text{ISC 2024}$
$\displaystyle \text{Answer the following questions.}$
$\displaystyle \text{(i) Find the equation of the plane containing the points } A,B \text{ and } C.$
$\displaystyle \text{(ii) Find the equation of the line } PQ.$
$\displaystyle \text{(iii) Calculate the height of the tower.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given points are } A(1,0,2),\; B(3,-1,1) \text{ and } C(1,2,1)$
$\displaystyle \text{Equation of plane is }$
$\displaystyle \begin{vmatrix} x-1 & y-0 & z-2\\ 3-1 & -1-0 & 1-2\\ 1-1 & 2-0 & 1-2 \end{vmatrix}=0$
$\displaystyle \Rightarrow \begin{vmatrix} x-1 & y & z-2\\ 2 & -1 & -1\\ 0 & 2 & -1 \end{vmatrix}=0$
$\displaystyle \Rightarrow (x-1)\begin{vmatrix}-1 & -1\\2 & -1\end{vmatrix} -y\begin{vmatrix}2 & -1\\0 & -1\end{vmatrix} +(z-2)\begin{vmatrix}2 & -1\\0 & 2\end{vmatrix}=0$
$\displaystyle \Rightarrow (x-1)(1+2)-y(-2)+(z-2)(4)=0$
$\displaystyle \Rightarrow 3(x-1)+2y+4(z-2)=0$
$\displaystyle \Rightarrow 3x+2y+4z-11=0$
$\displaystyle \text{Answer: } 3x+2y+4z-11=0$
$\displaystyle \text{(ii) The point } P(2,3,1) \text{ and plane through } A,B,C \text{ is } 3x+2y+4z=11$
$\displaystyle \text{Thus, equation of line } PQ \text{ is }$
$\displaystyle \frac{x-2}{3}=\frac{y-3}{2}=\frac{z-1}{4}$
$\displaystyle \text{Answer: } \frac{x-2}{3}=\frac{y-3}{2}=\frac{z-1}{4}$
$\displaystyle \text{ (iii) Equation of plane is } 3x+2y+4z=11$
$\displaystyle \text{Height of tower } P(2,3,1) \text{ is perpendicular distance from plane}$
$\displaystyle =\frac{|3(2)+2(3)+4(1)-11|}{\sqrt{9+4+16}}$
$\displaystyle =\frac{|6+6+4-11|}{\sqrt{29}}=\frac{5}{\sqrt{29}}$
$\displaystyle \text{Answer: } \frac{5}{\sqrt{29}}$
$\\$

$\displaystyle \textbf{Question 4: } \text{Find the equation of the plane passing through the } \text{point } (1,1,-1)$
$\displaystyle \text{ and perpendicular to the planes } x+2y+3z=7 \text{ and } 2x-3y+4z=0. \quad \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let equation of plane passing through } (1,1,-1) \text{ be }$
$\displaystyle a(x-1)+b(y-1)+c(z+1)=0 \quad ...(i)$
$\displaystyle \text{Since plane is perpendicular to } x+2y+3z=7 \text{ and } 2x-3y+4z=0$
$\displaystyle \therefore a+2b+3c=0 \quad ...(ii)$
$\displaystyle \text{and } 2a-3b+4c=0 \quad ...(iii)$
$\displaystyle \text{Solving (ii) and (iii) by cross-multiplication,}$
$\displaystyle \frac{a}{8+9}=\frac{b}{6-4}=\frac{c}{-3-4}$
$\displaystyle \Rightarrow \frac{a}{17}=\frac{b}{2}=\frac{c}{-7}=\lambda$
$\displaystyle \Rightarrow a=17\lambda,\; b=2\lambda,\; c=-7\lambda$
$\displaystyle \text{Substituting in (i),}$
$\displaystyle 17(x-1)+2(y-1)-7(z+1)=0$
$\displaystyle \Rightarrow 17x-17+2y-2-7z-7=0$
$\displaystyle \Rightarrow 17x+2y-7z=26$
$\displaystyle \text{Answer: } 17x+2y-7z=26$
$\\$

$\displaystyle \textbf{Question 5: } \text{A line passes through the point } (2,-1,3) \text{ and is perpendicular to the lines}$
$\displaystyle \overrightarrow{r}=(\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}) +\lambda(2\overrightarrow{i}-2\overrightarrow{j}+\overrightarrow{k}) \text{ and}$
$\displaystyle \overrightarrow{r}=(2\overrightarrow{i}-\overrightarrow{j}-3\overrightarrow{k}) +\mu(\overrightarrow{i}+2\overrightarrow{j}+2\overrightarrow{k}). \text{ Obtain its equation.} \quad \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equation of line passing through } (2,-1,3) \text{ be } \overrightarrow{r}=(2\hat{i}-\hat{j}+3\hat{k})+t\overrightarrow{b}$
$\displaystyle \text{Since, this line is perpendicular to the lines }$
$\displaystyle \overrightarrow{r}=\hat{i}+\hat{j}-\hat{k}+\lambda(2\hat{i}-2\hat{j}+\hat{k})$
$\displaystyle \text{and } \overrightarrow{r}=2\hat{i}-\hat{j}-3\hat{k}+\mu(\hat{i}+2\hat{j}+2\hat{k})$
$\displaystyle \text{Direction ratios are } (2,-2,1) \text{ and } (1,2,2)$
$\displaystyle \therefore \overrightarrow{b}=(2\hat{i}-2\hat{j}+\hat{k})\times(\hat{i}+2\hat{j}+2\hat{k})$
$\displaystyle =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-2&1\\1&2&2\end{vmatrix}$
$\displaystyle =\hat{i}((-2)(2)-(1)(2))-\hat{j}((2)(2)-(1)(1))+\hat{k}((2)(2)-(-2)(1))$
$\displaystyle =\hat{i}(-4-2)-\hat{j}(4-1)+\hat{k}(4+2)$
$\displaystyle =-6\hat{i}-3\hat{j}+6\hat{k}=-3(2\hat{i}+\hat{j}-2\hat{k})$
$\displaystyle \text{Hence, equation of line is } \overrightarrow{r}=(2\hat{i}-\hat{j}+3\hat{k})+t(2\hat{i}+\hat{j}-2\hat{k})$
$\displaystyle \text{Answer: } \overrightarrow{r}=(2\hat{i}-\hat{j}+3\hat{k})+t(2\hat{i}+\hat{j}-2\hat{k})$
$\\$

$\displaystyle \textbf{Question 6: } \text{The distance of the point } 2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k} \text{ from the plane}$
$\displaystyle \overrightarrow{r}\cdot(\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k})=9 \text{ will be} \quad \text{ISC 2023}$
$\displaystyle \text{(a) } 13 \quad \text{(b) } \frac{13}{\sqrt{21}} \quad \text{(c) } 21 \quad \text{(d) } \frac{21}{\sqrt{13}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) We know distance of a point } \overrightarrow{a} \text{ from plane } \overrightarrow{r}\cdot\overrightarrow{n}=d \text{ is }$
$\displaystyle \frac{|\overrightarrow{a}\cdot\overrightarrow{n}-d|} {|\overrightarrow{n}|}$
$\displaystyle \text{Here, } \overrightarrow{a}=2\hat{i}+\hat{j}-\hat{k},\; \overrightarrow{n}=\hat{i}-2\hat{j}+4\hat{k},\; d=-9$
$\displaystyle \Rightarrow \text{Distance}= \frac{|(2\hat{i}+\hat{j}-\hat{k})\cdot(\hat{i}-2\hat{j}+4\hat{k})-(-9)|} {\sqrt{1+4+16}}$
$\displaystyle =\frac{|2-2-4+9|}{\sqrt{21}}$
$\displaystyle =\frac{13}{\sqrt{21}}$
$\displaystyle \text{Answer: Distance }=\frac{13}{\sqrt{21}}$
$\\$

$\displaystyle \textbf{Question 7: } \text{Write the equation of the plane passing through the } \text{point } (2,4,6)$
$\displaystyle \text{ and making equal intercepts on the coordinate axes.} \quad \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Equation of plane in intercept form is } \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
$\displaystyle \text{Here, } a=b=c$
$\displaystyle \Rightarrow x+y+z=a$
$\displaystyle \text{Since plane passes through } (2,4,6)$
$\displaystyle \Rightarrow 2+4+6=a=12$
$\displaystyle \text{Required equation is } x+y+z=12$
$\displaystyle \text{Answer: } x+y+z=12$
$\\$

$\displaystyle \textbf{Question 8: } \text{Find the angle between the two lines}$
$\displaystyle \frac{x+1}{2}=\frac{y-2}{5}=\frac{z+3}{4} \text{ and } \frac{x-1}{5}=\frac{y+2}{2}=\frac{z-1}{-5}. \quad \text{ISC 2020}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given lines }\frac{x+1}{2}=\frac{y-2}{5}=\frac{z+3}{4},\ \frac{x-1}{5}=\frac{y+2}{2}=\frac{z-1}{-5}$
$\displaystyle \text{Direction ratios: }(2,5,4)\text{ and }(5,2,-5)$
$\displaystyle \cos\theta=\frac{(2)(5)+(5)(2)+(4)(-5)}{\sqrt{2^{2}+5^{2}+4^{2}}\sqrt{5^{2}+2^{2}+(-5)^{2}}}$
$\displaystyle =\frac{10+10-20}{\sqrt{4+25+16}\sqrt{25+4+25}}=0$
$\displaystyle \therefore \theta=\frac{\pi}{2}$
$\\$

$\displaystyle \textbf{Question 9: } \text{Determine the equation of the line passing through the point}$
$\displaystyle (-1,-3,-2) \text{ and perpendicular to the lines} \frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text{ and } \\c\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}. \quad \text{ISC 2020}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equation of line passing through the point } (-1,-3,-2) \text{ be } \frac{x+1}{a}=\frac{y+3}{b}=\frac{z+2}{c}$
$\displaystyle \text{Since, the above line is perpendicular to the lines } \frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text{ and } \frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}$
$\displaystyle \therefore a+2b+3c=0 \quad ...(i)$
$\displaystyle \text{and } -3a+2b+5c=0 \quad ...(ii)$
$\displaystyle \text{By cross-multiplication, we have } \frac{a}{10-6}=\frac{-b}{5+9}=\frac{c}{2+6}$
$\displaystyle \Rightarrow \frac{a}{4}=\frac{b}{-14}=\frac{c}{8}$
$\displaystyle \Rightarrow \frac{a}{2}=\frac{b}{-7}=\frac{c}{4}=\lambda$
$\displaystyle \Rightarrow a=2\lambda,\; b=-7\lambda,\; c=4\lambda$
$\displaystyle \therefore \text{Equation of line is } \frac{x+1}{2}=\frac{y+3}{-7}=\frac{z+2}{4}$
$\displaystyle \text{Answer: } \frac{x+1}{2}=\frac{y+3}{-7}=\frac{z+2}{4}$
$\\$

$\displaystyle \textbf{Question 10: } \text{Find the equation of the plane passing through the } \text{point } (2,-3,1)$
$\displaystyle \text{ and perpendicular to the line joining } \text{the points } (4,5,0) \text{ and } (1,-2,4). \quad \text{ISC 2020}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, direction ratios of normal to the line joining } (4,5,0) \text{ and } (1,-2,4) \text{ is } (1-4,-2-5,4-0)=(-3,-7,4)$
$\displaystyle \text{Equation of plane is }$
$\displaystyle -3(x-2)-7(y+3)+4(z-1)=0$
$\displaystyle \Rightarrow -3x+6-7y-21+4z-4=0$
$\displaystyle \Rightarrow -3x-7y+4z-19=0$
$\displaystyle \Rightarrow 3x+7y-4z+19=0$
$\displaystyle \text{Answer: } 3x+7y-4z+19=0$
$\\$

$\displaystyle \textbf{Question 11: } \text{Find the angle between the two lines } 2x=3y=-z \text{ and }$
$\displaystyle 6x=-y=-4z. \quad \text{ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equations of lines are written as } \frac{x}{1}=\frac{y}{2}=\frac{z}{-1} \text{ and } \frac{x}{6}=\frac{y}{-1}=\frac{z}{4}$
$\displaystyle \text{Here, direction ratios of above lines are } \langle a_1,b_1,c_1\rangle=\left\langle 1,2,-1\right\rangle$
$\displaystyle \text{and } \langle a_2,b_2,c_2\rangle=\left\langle 6,-1,4\right\rangle$
$\displaystyle \therefore \text{Angle between two lines is } \cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\displaystyle =\frac{(1)(6)+(2)(-1)+(-1)(4)}{\sqrt{1+4+1}\sqrt{36+1+16}}$
$\displaystyle =\frac{6-2-4}{\sqrt{6}\sqrt{53}}=0$
$\displaystyle \Rightarrow \theta=90^\circ$
$\displaystyle \text{Answer: The given lines are perpendicular.}$
$\\$

$\displaystyle \textbf{Question 12: } \text{Find the length of the perpendicular from the origin } \text{to the plane }$
$\displaystyle \overrightarrow{r}\cdot(3\overrightarrow{i}-4\overrightarrow{j}-12\overrightarrow{k})+39=0. \quad \text{ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the equation of plane which is at a distance } d \text{ from origin and having unit vector } \overrightarrow{n} \text{ is }$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=d \quad ...(i)$
$\displaystyle \text{Given, equation of plane can be written as }$
$\displaystyle \overrightarrow{r}\cdot(3\hat{i}-4\hat{j}+12\hat{k})=-39$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(-3\hat{i}+4\hat{j}-12\hat{k})=39 \quad ...(ii)$
$\displaystyle \text{Now, } \sqrt{(-3)^2+4^2+(-12)^2}=\sqrt{9+16+144}=13$
$\displaystyle \text{Dividing both sides of (ii) by } 13$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot\left(-\frac{3}{13}\hat{i}+\frac{4}{13}\hat{j}-\frac{12}{13}\hat{k}\right)=3 \quad ...(iii)$
$\displaystyle \text{Comparing (i) and (iii), we get } d=3$
$\displaystyle \text{Answer: } 3$
$\\$

$\displaystyle \textbf{Question 13: } \text{Find the image of a point having position vector}$
$\displaystyle 3\overrightarrow{i}-2\overrightarrow{j}+\overrightarrow{k} \text{ in the plane} \overrightarrow{r}\cdot(3\overrightarrow{i}-\overrightarrow{j}+4\overrightarrow{k})=2. \quad \text{ISC 2018, 17}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let point } Q(\alpha,\beta,\gamma) \text{ be the image of point } P(3,-2,1)$
$\displaystyle \text{in the plane } \overrightarrow{r}\cdot(3\hat{i}-\hat{j}+4\hat{k})=2$
$\displaystyle \text{Direction ratios of normal to the plane are } (3,-1,4)$
$\displaystyle \text{and direction ratios of line } PQ \text{ are } (\alpha-3,\beta+2,\gamma-1)$
$\displaystyle \text{Line } PQ \text{ is parallel to normal of plane.}$
$\displaystyle \therefore \frac{\alpha-3}{3}=\frac{\beta+2}{-1}=\frac{\gamma-1}{4}=\lambda$
$\displaystyle \Rightarrow \alpha=3\lambda+3,\; \beta=-\lambda-2,\; \gamma=4\lambda+1$
$\displaystyle R \text{ is mid-point of } PQ$
$\displaystyle \therefore R\left(\frac{3\lambda+6}{2},\frac{-\lambda-4}{2},\frac{4\lambda+2}{2}\right)$
$\displaystyle \text{Since, } R \text{ lies on plane } 3x-y+4z=2$
$\displaystyle \Rightarrow 3\left(\frac{3\lambda+6}{2}\right)-\left(\frac{-\lambda-4}{2}\right)+4\left(\frac{4\lambda+2}{2}\right)=2$
$\displaystyle \Rightarrow \frac{9\lambda+18+\lambda+4+16\lambda+8}{2}=2$
$\displaystyle \Rightarrow 26\lambda+30=4 \Rightarrow \lambda=-1$
$\displaystyle \therefore Q(0,-1,-3)$
$\displaystyle \text{Image of } P(3,-2,1) \text{ in the plane is } (0,-1,-3)$
$\displaystyle \text{Answer: } (0,-1,-3)$
$\\$

$\displaystyle \textbf{Question 14: } \text{The cartesian equation of a line is } 2x-3=3y+1=5-6z.$
$\displaystyle \text{Find the vector equation of a } \text{line passing through } (7,-5,0) \text{ and parallel to the}$
$\displaystyle \text{given line.} \quad \text{ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, cartesian equation of line is } 2x-3=3y+1=5-6z$
$\displaystyle \Rightarrow \frac{x-\frac{3}{2}}{\frac{1}{2}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-\frac{5}{6}}{-\frac{1}{6}}$
$\displaystyle \Rightarrow \frac{x-\frac{3}{2}}{1}=\frac{y+\frac{1}{3}}{1}=\frac{z-\frac{5}{6}}{-1}$
$\displaystyle \text{Direction ratios of line are } (1,1,-1)$
$\displaystyle \text{Vector equation of line passing through } (7,-5,0) \text{ and parallel to given line is }$
$\displaystyle \overrightarrow{r}=(7\hat{i}-5\hat{j})+\lambda(\hat{i}+\hat{j}-\hat{k})$
$\displaystyle \text{Answer: } \overrightarrow{r}=(7\hat{i}-5\hat{j})+\lambda(\hat{i}+\hat{j}-\hat{k})$
$\\$

$\displaystyle \textbf{Question 15: } \text{Show that the lines } \frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7} \text{ and}$
$\displaystyle \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8} \text{ intersect. Find the coordinates of } \text{their point of} \\ \text{intersection.} \text{ISC 2017}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given lines are } \frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}=\lambda \quad ...(i)$
$\displaystyle \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}=\mu \quad ...(ii)$
$\displaystyle \text{General point on line (i) is } M(\lambda+4,-4\lambda-3,7\lambda-1)$
$\displaystyle \text{General point on line (ii) is } N(2\mu+1,-3\mu-1,8\mu-10)$
$\displaystyle \text{The given lines will intersect only when } M \text{ and } N \text{ coincide for some } \lambda \text{ and } \mu$
$\displaystyle \text{Given, lines will intersect only when } \lambda+4=2\mu+1, -4\lambda-3=-3\mu-1 \text{ and } 7\lambda-1=8\mu-10$
$\displaystyle \Rightarrow \lambda-2\mu=-3 \quad ...(iii)$
$\displaystyle \Rightarrow -4\lambda+3\mu=2 \quad ...(iv)$
$\displaystyle \Rightarrow 7\lambda-8\mu=-9 \quad ...(v)$
$\displaystyle \text{On solving (iii) and (iv), we get } \lambda=1,\; \mu=2$
$\displaystyle \text{These values also satisfy Eq. (v)}$
$\displaystyle \text{Hence, point of intersection is } (5,-7,6)$
$\displaystyle \text{Answer: } (5,-7,6)$
$\\$

$\displaystyle \textbf{Question 16: } \text{Find the equation of the plane passing through the \text{point } (1,-2,1) }$
$\displaystyle \text{ and perpendicular to the line joining } \text{points } A(3,2,1) \text{ and } B(1,4,2). \quad \text{ISC 2017}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since, the plane passes through the point } A(1,-2,1)$
$\displaystyle \text{Position vector of } A \text{ is } \hat{i}-2\hat{j}+\hat{k}$
$\displaystyle \text{Normal vector } \overrightarrow{n} \text{ is parallel to line joining } A(3,2,1) \text{ and } B(1,4,2)$
$\displaystyle \therefore \overrightarrow{n}=\lambda(-2\hat{i}+2\hat{j}+\hat{k})$
$\displaystyle \text{Vector equation of plane is } \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(-2\hat{i}+2\hat{j}+\hat{k})=(\hat{i}-2\hat{j}+\hat{k})\cdot(-2\hat{i}+2\hat{j}+\hat{k})$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(-2\hat{i}+2\hat{j}+\hat{k})=-2-4+1=-5$
$\displaystyle \Rightarrow -2x+2y+z=-5$
$\displaystyle \text{Answer: } -2x+2y+z=-5$
$\\$

$\displaystyle \textbf{Question 17: } \text{Find the image of the point } (2,-1,5) \text{ in the line}$
$\displaystyle \frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}. \text{ Also, find the length of the } \text{perpendicular from the point }$
$\displaystyle (2,-1,5) \text{ to the line.} \quad \text{ISC 2016}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, line is } \frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11} \quad ...(i)$
$\displaystyle \text{Let } N \text{ be the foot of perpendicular from } P(2,-1,5)$
$\displaystyle \text{Any point on line (i) is } (10\lambda+11,-4\lambda-2,-11\lambda-8)$
$\displaystyle \text{Direction ratios of } NP \text{ are } \langle 10\lambda+9,-4\lambda-1,-11\lambda-13\rangle$
$\displaystyle \text{Direction ratios of line are } \langle 10,-4,-11\rangle$
$\displaystyle \therefore 10(10\lambda+9)-4(-4\lambda-1)-11(-11\lambda-13)=0$
$\displaystyle \Rightarrow 100\lambda+90+16\lambda+4+121\lambda+143=0$
$\displaystyle \Rightarrow 237\lambda+237=0 \Rightarrow \lambda=-1$
$\displaystyle \text{Coordinates of } N=(1,2,3)$
$\displaystyle \text{Let image of } P(2,-1,5) \text{ be } P'(\alpha,\beta,\gamma)$
$\displaystyle \text{Since } N \text{ is midpoint of } PP'$
$\displaystyle 1=\frac{2+\alpha}{2},\; 2=\frac{-1+\beta}{2},\; 3=\frac{5+\gamma}{2}$
$\displaystyle \Rightarrow \alpha=0,\; \beta=5,\; \gamma=1$
$\displaystyle \text{Hence image is } P'(0,5,1)$
$\displaystyle \text{Required perpendicular length is } \sqrt{(2-1)^2+(-1-2)^2+(5-3)^2}$
$\displaystyle =\sqrt{1+9+4}=\sqrt{14}$
$\displaystyle \text{Answer: Length }=\sqrt{14}$
$\\$

$\displaystyle \textbf{Question 18: } \text{Find the shortest distance between the lines}$
$\displaystyle \overrightarrow{r}_{1}=\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k} +\lambda(2\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k}) \text{ and}$
$\displaystyle \overrightarrow{r}_{2}=2\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k} +\mu(4\overrightarrow{i}+6\overrightarrow{j}+8\overrightarrow{k}). \quad \text{ISC 2015}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, equations of lines are }$
$\displaystyle \overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k} +\lambda(2\hat{i}+3\hat{j}+4\hat{k})$
$\displaystyle \text{and } \overrightarrow{r}=2\hat{i}+4\hat{j}+5\hat{k} +\mu(4\hat{i}+6\hat{j}+8\hat{k})$
$\displaystyle \text{Comparing with } \overrightarrow{r}=\overrightarrow{a} +\lambda\overrightarrow{b}$
$\displaystyle \overrightarrow{a_1}=\hat{i}+2\hat{j}+3\hat{k},\; \overrightarrow{b_1}=2\hat{i}+3\hat{j}+4\hat{k}$
$\displaystyle \overrightarrow{a_2}=2\hat{i}+4\hat{j}+5\hat{k},\; \overrightarrow{b_2}=4\hat{i}+6\hat{j}+8\hat{k}$
$\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1} =(\hat{i}+2\hat{j}+2\hat{k})$
$\displaystyle \overrightarrow{b_1}\times\overrightarrow{b_2} =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\4&6&8\end{vmatrix}$
$\displaystyle =\hat{i}(24-24)-\hat{j}(16-16)+\hat{k}(12-12)$
$\displaystyle =0\hat{i}-0\hat{j}+0\hat{k}$
$\displaystyle \text{Hence, lines are parallel}$
$\displaystyle \text{Shortest distance }= \frac{|\overrightarrow{b_1}\times(\overrightarrow{a_2}-\overrightarrow{a_1})|} {|\overrightarrow{b_1}|}$
$\displaystyle =\frac{\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\1&2&2\end{vmatrix}} {\sqrt{4+9+16}}$
$\displaystyle =\frac{\hat{i}(6-8)-\hat{j}(4-4)+\hat{k}(4-3)}{\sqrt{29}}$
$\displaystyle =\frac{-2\hat{i}+0\hat{j}+\hat{k}}{\sqrt{29}}$
$\displaystyle =\frac{\sqrt{4+1}}{\sqrt{29}}=\sqrt{\frac{5}{29}}$
$\displaystyle \text{Answer: Shortest distance }=\sqrt{\frac{5}{29}}$
$\\$

$\displaystyle \textbf{Question 19: } \text{Find the equation of a line passing through the points } P(-1,3,2)$
$\displaystyle \text{ and } Q(-4,2,-2). \text{ Also, if the point } R(5,5,\lambda) \text{ is collinear with the points } P \text{ and } Q, \text{ then}$
$\displaystyle \text{find the value of } \lambda. \quad \text{ISC 2014}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equation of line passing through } P(-1,3,2) \text{ and } Q(-4,2,-2) \text{ be }$
$\displaystyle \frac{x+1}{-4+1}=\frac{y-3}{2-3}=\frac{z-2}{-2-2}$
$\displaystyle \left[\because \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\right]$
$\displaystyle \Rightarrow \frac{x+1}{-3}=\frac{y-3}{-1}=\frac{z-2}{-4}$
$\displaystyle \Rightarrow \frac{x+1}{3}=\frac{y-3}{1}=\frac{z-2}{4}=r \quad [\text{multiplying by } -1]$
$\displaystyle \therefore \text{Any point on the above line } PQ \text{ is } (3r-1,r+3,4r+2)$
$\displaystyle \text{Given, the point } R(5,5,\lambda) \text{ lies on the line } PQ$
$\displaystyle \therefore (3r-1,r+3,4r+2)=(5,5,\lambda)$
$\displaystyle \Rightarrow 3r-1=5$
$\displaystyle \Rightarrow 3r=6$
$\displaystyle \Rightarrow r=2$
$\displaystyle \text{and } 4r+2=\lambda$
$\displaystyle \Rightarrow 4\times 2+2=\lambda$
$\displaystyle \Rightarrow 10=\lambda$
$\displaystyle \text{Hence, the value of } \lambda \text{ is } 10$
$\displaystyle \text{Answer: } \lambda=10$
$\\$

$\displaystyle \textbf{Question 20: } \text{Find the equation of the plane passing through the } \text{points } (2,-3,1) \text{ and }$
$\displaystyle (-1,1,-7) \text{ and perpendicular to } \text{the plane } x-2y+5z+1=0. \quad \text{ISC 2014}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Equation of plane passing through } (2,-3,1) \text{ is } a(x-2)+b(y+3)+c(z-1)=0 \quad ...(i)$
$\displaystyle \text{Also passes through } (-1,1,-7)$
$\displaystyle \Rightarrow a(-3)+b(4)+c(-8)=0$
$\displaystyle \Rightarrow -3a+4b-8c=0 \Rightarrow 3a-4b+8c=0 \quad ...(ii)$
$\displaystyle \text{Also perpendicular to plane } x-2y+5z+1=0$
$\displaystyle \Rightarrow a-2b+5c=0 \quad ...(iii)$
$\displaystyle \text{Solving (ii) and (iii),}$
$\displaystyle \frac{a}{-20+16}=\frac{b}{8-15}=\frac{c}{-6+4}$
$\displaystyle \Rightarrow \frac{a}{-4}=\frac{b}{-7}=\frac{c}{-2}$
$\displaystyle \Rightarrow \frac{a}{4}=\frac{b}{7}=\frac{c}{2}$
$\displaystyle \Rightarrow a=2c,\; b=\frac{7c}{2}$
$\displaystyle \text{Substitute in (i): } 2c(x-2)+\frac{7c}{2}(y+3)+c(z-1)=0$
$\displaystyle \Rightarrow 4(x-2)+7(y+3)+2(z-1)=0$
$\displaystyle \Rightarrow 4x-8+7y+21+2z-2=0$
$\displaystyle \Rightarrow 4x+7y+2z+11=0$
$\displaystyle \text{Answer: } 4x+7y+2z+11=0$
$\\$

$\displaystyle \textbf{Question 21: } \text{Find the equation of the plane passing through the } \text{points } (2,-3,1) \text{ and }$
$\displaystyle (-1,1,-7) \text{ and perpendicular to} \text{the plane } x-2y+5z+1=0. \quad \text{ISC 2014}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Equation of plane passing through } (2,-3,1) \text{ is } a(x-2)+b(y+3)+c(z-1)=0 \quad ...(i)$
$\displaystyle \text{Also passes through } (-1,1,-7)$
$\displaystyle \Rightarrow a(-3)+b(4)+c(-8)=0$
$\displaystyle \Rightarrow -3a+4b-8c=0 \Rightarrow 3a-4b+8c=0 \quad ...(ii)$
$\displaystyle \text{Also perpendicular to plane } x-2y+5z+1=0$
$\displaystyle \Rightarrow a-2b+5c=0 \quad ...(iii)$
$\displaystyle \text{Solving (ii) and (iii),}$
$\displaystyle \frac{a}{-20+16}=\frac{b}{8-15}=\frac{c}{-6+4}$
$\displaystyle \Rightarrow \frac{a}{-4}=\frac{b}{-7}=\frac{c}{-2}$
$\displaystyle \Rightarrow \frac{a}{4}=\frac{b}{7}=\frac{c}{2}$
$\displaystyle \Rightarrow a=2c,\; b=\frac{7c}{2}$
$\displaystyle \text{Substitute in (i): } 2c(x-2)+\frac{7c}{2}(y+3)+c(z-1)=0$
$\displaystyle \Rightarrow 4(x-2)+7(y+3)+2(z-1)=0$
$\displaystyle \Rightarrow 4x-8+7y+21+2z-2=0$
$\displaystyle \Rightarrow 4x+7y+2z+11=0$
$\displaystyle \text{Answer: } 4x+7y+2z+11=0$
$\\$

$\displaystyle \textbf{Question 22: } \text{Find the equation of the planes parallel to the plane}$
$\displaystyle 2x-4y+4z=7 \text{ and which are at a distance of five } \text{units from the point } (3,-1,2). \quad \text{ISC 2013}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Equation of plane parallel to } 2x-4y+4z=7 \text{ is }$
$\displaystyle 2x-4y+4z+k=0 \quad ...(i)$
$\displaystyle \text{Distance from } (3,-1,2) \text{ is } 5$
$\displaystyle \Rightarrow \frac{|2(3)-4(-1)+4(2)+k|}{\sqrt{4+16+16}}=5$
$\displaystyle \Rightarrow \frac{|6+4+8+k|}{6}=5$
$\displaystyle \Rightarrow |18+k|=30$
$\displaystyle \Rightarrow 18+k=30 \text{ or } 18+k=-30$
$\displaystyle \Rightarrow k=12 \text{ or } k=-48$
$\displaystyle \text{Hence equations are } 2x-4y+4z+12=0 \text{ and } 2x-4y+4z-48=0$
$\displaystyle \text{Answer: } 2x-4y+4z+12=0,\; 2x-4y+4z-48=0$
$\\$

$\displaystyle \textbf{Question 23: } \text{Find the equation of a line passing through the point } (-1,3,-2)$
$\displaystyle \text{ and perpendicular to the lines } \frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text{ and } \frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}. \quad \text{ISC 2013}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the direction ratios of the required line be } a,b \text{ and } c$
$\displaystyle \text{Since, the required line is perpendicular to the given lines.}$
$\displaystyle \text{Then, } a+2b+3c=0 \quad ...(i)$
$\displaystyle \left[\because a_1a_2+b_1b_2+c_1c_2=0\right]$
$\displaystyle \text{and } -3a+2b+5c=0 \quad ...(ii)$
$\displaystyle \text{On solving Eqs. (i) and (ii), by cross-multiplication method, we get}$
$\displaystyle \frac{a}{10-6}=\frac{b}{-9-5}=\frac{c}{2+6}$
$\displaystyle \Rightarrow \frac{a}{4}=\frac{b}{-14}=\frac{c}{8}$
$\displaystyle \Rightarrow \frac{a}{2}=\frac{b}{-7}=\frac{c}{4}=r \quad (\text{say})$
$\displaystyle \text{As, the required line passes through } (-1,3,-2) \text{ and has direction ratios}$
$\displaystyle \text{proportional to } (2,-7,4)$
$\displaystyle \text{Hence, the required equation of line is }$
$\displaystyle \frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}$
$\displaystyle \text{Answer: } \frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}$
$\\$

$\displaystyle \textbf{Question 24: } \text{Find the shortest distance between the lines}$
$\displaystyle \frac{x-8}{3}=\frac{y+9}{-16}=\frac{z-10}{7} \text{and } \frac{x-15}{3}=\frac{y-29}{8}=\frac{5-z}{5}. \quad \text{ISC 2012}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, lines are } \frac{x-8}{3}=\frac{y-(-9)}{-16}=\frac{z-10}{7}$
$\displaystyle \text{and } \frac{x-15}{3}=\frac{y-29}{8}=\frac{5-z}{5}$
$\displaystyle \text{or } \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$
$\displaystyle \text{On comparing with standard equation of line }$
$\displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}, \text{ we get}$
$\displaystyle x_1=8,\; y_1=-9,\; z_1=10,\; a_1=3,\; b_1=-16,\; c_1=7$
$\displaystyle x_2=15,\; y_2=29,\; z_2=5,\; a_2=3,\; b_2=8,\; c_2=-5$
$\displaystyle \therefore \text{The shortest distance between the lines}$
$\displaystyle = \frac{\left|\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1\\ a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \end{vmatrix}\right|} {\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}}$
$\displaystyle = \frac{\left|\begin{vmatrix} 15-8 & 29+9 & 5-10\\ 3 & -16 & 7\\ 3 & 8 & -5 \end{vmatrix}\right|} {\sqrt{((-16)(-5)-8(7))^2+(7(3)+5(3))^2+(3(8)-3(-16))^2}}$
$\displaystyle = \frac{\left|\begin{vmatrix} 7 & 38 & -5\\ 3 & -16 & 7\\ 3 & 8 & -5 \end{vmatrix}\right|} {\sqrt{(80-56)^2+(21+15)^2+(24+48)^2}}$
$\displaystyle = \frac{\left|7(80-56)-38(-15-21)-5(24+48)\right|} {\sqrt{(24)^2+(36)^2+(72)^2}}$
$\displaystyle = \frac{7(24)-38(-36)-5(72)} {\sqrt{12^2(2^2+3^2+6^2)}}$
$\displaystyle =\frac{168+1368-360}{12\sqrt{4+9+36}}$
$\displaystyle =\frac{1176}{12\sqrt{49}}$
$\displaystyle =\frac{98}{7}=14 \text{ units}$
$\displaystyle \text{Answer: } 14 \text{ units}$
$\\$

$\displaystyle \textbf{Question 25: } \text{Find the equation of the plane passing through the } \text{points } A(2,1,-3),$
$\displaystyle B(-3,-2,1) \text{ and } C(2,4,-1). \quad \text{ISC 2011}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Equation of plane passing through } (2,1,-3),(-3,-2,1),(2,4,-1)$
$\displaystyle \text{is } \begin{vmatrix}x-2&y-1&z+3\\-5&-3&4\\0&3&2\end{vmatrix}=0$
$\displaystyle \Rightarrow (x-2)\begin{vmatrix}-3&4\\3&2\end{vmatrix}-(y-1)\begin{vmatrix}-5&4\\0&2\end{vmatrix}+(z+3)\begin{vmatrix}-5&-3\\0&3\end{vmatrix}=0$
$\displaystyle \Rightarrow (x-2)(-6-12)-(y-1)(-10)+(z+3)(-15)=0$
$\displaystyle \Rightarrow -18(x-2)+10(y-1)-15(z+3)=0$
$\displaystyle \Rightarrow -18x+36+10y-10-15z-45=0$
$\displaystyle \Rightarrow 18x-10y+15z+19=0$
$\displaystyle \text{Answer: } 18x-10y+15z+19=0$
$\\$

$\displaystyle \textbf{Question 26: } \text{Find the equation of the plane which contains the line}$
$\displaystyle \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{4} \text{ and is perpendicular to the plane } \\ x+2y+z=12. \quad \text{ISC 2009}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let required equation of plane be } ax+by+cz+d=0$
$\displaystyle \text{Given, equation of plane and line are } x+2y+z-12=0$
$\displaystyle \text{and } \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{4}$
$\displaystyle \text{Here, } (x_1,y_1,z_1)=(1,-1,3) \text{ and DR's are } (2,-1,4)$
$\displaystyle \text{Plane is perpendicular to line } \Rightarrow 2a-b+4c=0 \quad ...(ii)$
$\displaystyle \text{Also plane is perpendicular to plane } x+2y+z=12$
$\displaystyle \Rightarrow a+2b+c=0 \quad ...(iii)$
$\displaystyle \text{Solving (ii) and (iii)}$
$\displaystyle \frac{a}{-1-8}=\frac{b}{4-2}=\frac{c}{4+1}$
$\displaystyle \Rightarrow \frac{a}{-9}=\frac{b}{2}=\frac{c}{5}$
$\displaystyle \Rightarrow a=9k,\; b=-2k,\; c=-5k$
$\displaystyle \text{Substituting in } ax+by+cz+d=0$
$\displaystyle 9x-2y-5z+d=0$
$\displaystyle \text{Since plane passes through } (1,-1,3)$
$\displaystyle \Rightarrow 9(1)-2(-1)-5(3)+d=0$
$\displaystyle \Rightarrow 9+2-15+d=0 \Rightarrow d=4$
$\displaystyle \text{Hence equation is } 9x-2y-5z+4=0$
$\displaystyle \text{Answer: } 9x-2y-5z+4=0$
$\\$

$\displaystyle \textbf{Question 27: } \text{Find the coordinates of the point where the line } \text{joining the points }$
$\displaystyle (1,-2,3) \text{ and } (2,-1,5) \text{ cuts the plane } x-2y+3z=19.$
$\displaystyle \text{ Hence, find the distance of this point } \text{from the point } (5,4,1). \quad \text{ISC 2008}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Direction ratios of line joining } (1,-2,3) \text{ and } (2,-1,5)$
$\displaystyle \text{are } (1,1,2)$
$\displaystyle \text{Equation of line is } \frac{x-1}{1}=\frac{y+2}{1}=\frac{z-3}{2}=r$
$\displaystyle \text{Any point on line is } (1+r,-2+r,3+2r)$
$\displaystyle \text{This lies on plane } x-2y+3z=19$
$\displaystyle \Rightarrow 1+r-2(-2+r)+3(3+2r)=19$
$\displaystyle \Rightarrow 1+r+4-2r+9+6r=19$
$\displaystyle \Rightarrow 5r+14=19 \Rightarrow r=1$
$\displaystyle \therefore P=(2,-1,5)$
$\displaystyle \text{Distance between } (2,-1,5) \text{ and } (5,4,1)$
$\displaystyle =\sqrt{(5-2)^2+(4+1)^2+(1-5)^2}$
$\displaystyle =\sqrt{9+25+16}=\sqrt{50}=5\sqrt{2}$
$\displaystyle \text{Answer: } 5\sqrt{2}$
$\\$

$\displaystyle \textbf{Question 28: } \text{Find the equation of the plane through the point } (1,2,3) \text{ and}$
$\displaystyle \text{perpendicular to the planes } x+y+2z=3 \text{ and } 3x+2y+z=4. \quad \text{ISC 2007}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that equation of plane passing through } (x_1,y_1,z_1) \text{ is}$
$\displaystyle a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
$\displaystyle \text{Equation of plane passing through } (1,2,3) \text{ is}$
$\displaystyle a(x-1)+b(y-2)+c(z-3)=0 \quad ...(i)$
$\displaystyle \text{Since, plane is perpendicular to } x+y+2z=3 \text{ and } 3x+2y+z=4$
$\displaystyle \therefore a+b+2c=0 \quad ...(ii)$
$\displaystyle \text{and } 3a+2b+c=0 \quad ...(iii)$
$\displaystyle \text{On solving (ii) and (iii), we get}$
$\displaystyle \frac{a}{-4-1}=\frac{b}{6-1}=\frac{c}{2-3}$
$\displaystyle \Rightarrow \frac{a}{-5}=\frac{b}{5}=\frac{c}{-1}$
$\displaystyle \Rightarrow a=3c,\; b=-5c$
$\displaystyle \text{Putting in Eq. (i): } 3c(x-1)-5c(y-2)+c(z-3)=0$
$\displaystyle \Rightarrow c(3x-3-5y+10+z-3)=0$
$\displaystyle \Rightarrow 3x-5y+z+4=0$
$\displaystyle \text{Answer: } 3x-5y+z+4=0$
$\\$

$\displaystyle \textbf{Question 29: } \text{Find the equation of the two planes passing through the } \text{points } (0,4,-3)$
$\displaystyle \text{ and } (6,-4,3), \text{ if the sum of their } \text{intercepts on the three axes is zero.} \quad \text{ISC 2006}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Suppose the equation of the plane is } ax+by+cz+d=0 \quad ...(i)$
$\displaystyle \text{Since, plane passes through } (0,4,-3) \text{ and } (6,-4,3)$
$\displaystyle \Rightarrow 0+4b-3c+d=0 \quad ...(ii)$
$\displaystyle \text{and } 6a-4b+3c+d=0 \quad ...(iii)$
$\displaystyle \text{On adding (ii) and (iii), we get } 6a+2d=0$
$\displaystyle \Rightarrow a=-\frac{2d}{6}=-\frac{d}{3}$
$\displaystyle \text{On putting } a=-\frac{d}{3} \text{ in (iii), we get}$
$\displaystyle 6\left(-\frac{d}{3}\right)-4b+3c+d=0$
$\displaystyle \Rightarrow -2d-4b+3c+d=0$
$\displaystyle \Rightarrow -4b+3c-d=0$
$\displaystyle \Rightarrow 4b-3c=-d \Rightarrow b=\frac{3c-d}{4}$
$\displaystyle \text{Considering } ax+by+cz+d=0 \text{ in intercept form }$
$\displaystyle \frac{x}{\frac{-d}{a}}+\frac{y}{\frac{-d}{b}}+\frac{z}{\frac{-d}{c}}=1$
$\displaystyle \text{According to condition, sum of intercepts is zero}$
$\displaystyle \Rightarrow -\frac{d}{a}-\frac{d}{b}-\frac{d}{c}=0$
$\displaystyle \Rightarrow -d\left[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]=0$
$\displaystyle \Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
$\displaystyle \text{Substituting } a=-\frac{d}{3},\; b=\frac{3c-d}{4}$
$\displaystyle \Rightarrow \frac{-3}{d}+\frac{4}{3c-d}+\frac{1}{c}=0$
$\displaystyle \Rightarrow -\frac{3(3c-d)c+4dc+d(3c-d)}{d(3c-d)c}=0$
$\displaystyle \Rightarrow -9c^2+3cd+4cd+3cd-d^2=0$
$\displaystyle \Rightarrow -9c^2+10cd-d^2=0$
$\displaystyle \Rightarrow 9c^2-10cd+d^2=0$
$\displaystyle \Rightarrow 9c^2-9cd-cd+d^2=0$
$\displaystyle \Rightarrow 9c(c-d)-d(c-d)=0$
$\displaystyle \Rightarrow (9c-d)(c-d)=0$
$\displaystyle \Rightarrow c=\frac{d}{9} \text{ or } c=d$
$\displaystyle \text{When } c=\frac{d}{9},$
$\displaystyle b=\frac{3\left(\frac{d}{9}\right)-d}{4}=\frac{\frac{d}{3}-d}{4}=\frac{-2d}{12}=-\frac{d}{6}$
$\displaystyle \text{When } c=d,$
$\displaystyle b=\frac{3d-d}{4}=\frac{2d}{4}=\frac{d}{2}$
$\displaystyle \text{Hence, required equations of planes are}$
$\displaystyle -\frac{d}{3}x-\frac{d}{6}y+\frac{d}{9}z+d=0$
$\displaystyle \Rightarrow -d(6x+3y-2z-18)=0$
$\displaystyle \Rightarrow 6x+3y-2z-18=0$
$\displaystyle \text{and } -\frac{d}{3}x+\frac{d}{2}y+d z+d=0$
$\displaystyle \Rightarrow -d(2x-3y-6z-6)=0$
$\displaystyle \Rightarrow 2x-3y-6z-6=0$
$\displaystyle \text{Answer: } 6x+3y-2z-18=0,\; 2x-3y-6z-6=0$
$\\$

$\displaystyle \textbf{Question 30: } \text{A plane passes through the point } (4,2,4) \text{ and is } \text{perpendicular to the planes }$
$\displaystyle 2x+5y+4z+1=0 \text{ and} 4x+7y+6z+2=0. \text{ Then, find the equation of the plane.}$
$\displaystyle \quad \text{ISC 2005}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Equation of plane passing through } (4,2,4) \text{ is }$
$\displaystyle a(x-4)+b(y-2)+c(z-4)=0 \quad ...(i)$
$\displaystyle \text{Plane is perpendicular to } 2x+5y+4z+1=0 \text{ and } 4x+7y+6z+2=0$
$\displaystyle \therefore 2a+5b+4c=0 \quad ...(ii)$
$\displaystyle \text{and } 4a+7b+6c=0 \quad ...(iii)$
$\displaystyle \text{Multiply (ii) by 2: } 4a+10b+8c=0 \quad ...(iv)$
$\displaystyle \text{Subtract (iii) from (iv): } 3b+2c=0 \Rightarrow b=-\frac{2c}{3}$
$\displaystyle \text{Substitute in (ii): } 2a+5\left(-\frac{2c}{3}\right)+4c=0$
$\displaystyle \Rightarrow 2a-\frac{10c}{3}+4c=0$
$\displaystyle \Rightarrow 6a-10c+12c=0 \Rightarrow 6a+2c=0$
$\displaystyle \Rightarrow a=-\frac{c}{3}$
$\displaystyle \text{Substitute in (i): } -\frac{c}{3}(x-4)-\frac{2c}{3}(y-2)+c(z-4)=0$
$\displaystyle \Rightarrow -\frac{c}{3}[x-4+2(y-2)-3(z-4)]=0$
$\displaystyle \Rightarrow x+2y-3z+4=0$
$\displaystyle \text{Answer: } x+2y-3z+4=0$
$\\$

$\displaystyle \textbf{Question 31: } \text{Find the equation of plane through the point } (-1,-1,2)$
$\displaystyle \text{and perpendicular to the planes } 3x+2y-3z=1 \text{ and } 5x-4y+z=5. \quad \text{ISC 2004}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, plane passes through the point } (-1,-1,2).$
$\displaystyle \text{Therefore, equation of plane is } a(x+1)+b(y+1)+c(z-2)=0 \quad ...(i)$
$\displaystyle \text{As, the plane is perpendicular to the planes } 3x+2y-3z=1 \text{ and } 5x-4y+z=5$
$\displaystyle \therefore 3a+2b-3c=0 \quad ...(ii)$
$\displaystyle \text{and } 5a-4b+c=0 \quad ...(iii)$
$\displaystyle \text{On solving Eqs. (ii) and (iii), we get}$
$\displaystyle \frac{a}{2-12}=\frac{-b}{-15-3}=\frac{c}{-12-10}$
$\displaystyle \Rightarrow \frac{a}{-10}=\frac{-b}{-18}=\frac{c}{-22}$
$\displaystyle \Rightarrow \frac{a}{10}=\frac{b}{18}=\frac{c}{22}$
$\displaystyle \Rightarrow a=\frac{5c}{11},\; b=\frac{9c}{11}$
$\displaystyle \text{Putting } a=\frac{5c}{11} \text{ and } b=\frac{9c}{11} \text{ in Eq. (i), we get}$
$\displaystyle \frac{5c}{11}(x+1)+\frac{9c}{11}(y+1)+c(z-2)=0$
$\displaystyle \Rightarrow \frac{c}{11}(5x+5+9y+9+11z-22)=0$
$\displaystyle \Rightarrow 5x+9y+11z-8=0$
$\displaystyle \text{Answer: } 5x+9y+11z-8=0$
$\\$

$\displaystyle \textbf{Question 32: } \text{Find the angle between the line } \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2} \\ \text{ and the plane } x+y+2z=0. \quad \text{ISC 2004}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, equation of plane is } x+y+2z=0 \text{ and line is } \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$
$\displaystyle \text{The direction ratios of normal to plane are } (1,1,2)$
$\displaystyle \text{The direction ratios of line are } (3,2,-2)$
$\displaystyle \therefore \cos\theta=\frac{|aa_1+bb_1+cc_1|} {\sqrt{a^2+b^2+c^2}\sqrt{a_1^2+b_1^2+c_1^2}}$
$\displaystyle =\frac{|1\times3+1\times2+2\times(-2)|}{\sqrt{1^2+1^2+2^2}\sqrt{3^2+2^2+(-2)^2}}$
$\displaystyle =\frac{|3+2-4|}{\sqrt{6}\sqrt{17}}=\frac{1}{\sqrt{102}}$
$\displaystyle \text{Answer: } \cos\theta=\frac{1}{\sqrt{102}}$
$\\$

$\displaystyle \textbf{Question 33: } \text{Find the cosine of the angle between the planes}$
$\displaystyle x+2y-2z+6=0 \text{ and } 2x+2y+z+8=0. \quad \text{ISC 2003}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given planes are } x+2y-2z+6=0 \text{ and } 2x+2y+z+8=0$
$\displaystyle \text{Direction ratios of normals to these planes are } \langle 1,2,-2\rangle \text{ and } \langle 2,2,1\rangle$
$\displaystyle \text{The acute angle } \theta \text{ between these planes is given by}$
$\displaystyle \cos\theta=\frac{|a_1a_2+b_1b_2+c_1c_2|} {\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\displaystyle \therefore \cos\theta=\frac{|1(2)+2(2)+(-2)(1)|}{\sqrt{1^2+2^2+(-2)^2}\sqrt{2^2+2^2+1^2}}$
$\displaystyle =\frac{|2+4-2|}{\sqrt{9}\sqrt{9}}=\frac{4}{9}$
$\displaystyle \text{Answer: } \cos\theta=\frac{4}{9}$
$\\$

$\displaystyle \textbf{Question 34: } \text{Find the equation of the plane passing through } (1,2,3)$
$\displaystyle \text{and perpendicular to the straight line } \frac{x}{-2}=\frac{y}{-4}=\frac{z}{3}. \quad \text{ISC 2002}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let equation of plane passing through } (1,2,3) \text{ be}$
$\displaystyle a(x-1)+b(y-2)+c(z-3)=0 \quad ...(i)$
$\displaystyle \text{Given that, plane is perpendicular to the straight line } \frac{x}{-2}=\frac{y}{-4}=\frac{z}{3}$
$\displaystyle \therefore \text{normal to the plane is parallel to the line.}$
$\displaystyle \text{Direction ratios of normal to the plane are proportional to the line.}$
$\displaystyle \text{i.e. } \frac{a}{-2}=\frac{b}{-4}=\frac{c}{3}=k$
$\displaystyle \text{Putting values of } a,b,c \text{ in Eq. (i), we get}$
$\displaystyle -2k(x-1)-4k(y-2)+3k(z-3)=0$
$\displaystyle \Rightarrow -2x+2-4y+8+3z-9=0$
$\displaystyle \Rightarrow -2x-4y+3z+1=0$
$\displaystyle \Rightarrow 2x+4y-3z-1=0$
$\displaystyle \text{Answer: } 2x+4y-3z-1=0$
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