Class 12: The Plane – Board Problems

$\displaystyle \textbf{Question 1:}\quad \text{Find the equation of the plane passing through the point }(1,1,-1) \\ \text{and perpendicular to the planes }x+2y+3z-7=0\text{ and }2x-3y+4z=0.\quad [\text{CBSE 2003}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{The equation of any plane passing through }(1,1,-1)\text{ is}$
$\displaystyle a(x-1)+b(y-1)+c(z+1)=0\qquad \ldots(i)$
$\displaystyle \text{If plane is perpendicular to each one of the planes }x+2y+3z-7=0\text{ and }2x-3y+4z=0,\text{ then}$
$\displaystyle a+2b+3c=0\qquad \ldots(ii)$
$\displaystyle 2a-3b+4c=0\qquad \ldots(iii)$
$\displaystyle \text{On solving (ii) and (iii) by cross-multiplication, we get}$
$\displaystyle \frac{a}{(2)(4)-(-3)(3)}=\frac{b}{(3)(2)-(1)(4)}=\frac{c}{(1)(-3)-(2)(2)}$
$\displaystyle \Rightarrow \frac{a}{17}=\frac{b}{2}=\frac{c}{-7}=\lambda\ (\text{say})$
$\displaystyle \Rightarrow a=17\lambda,\ b=2\lambda\text{ and }c=-7\lambda$
$\displaystyle \text{Putting }a=17\lambda,\ b=2\lambda\text{ and }c=-7\lambda\text{ in (i), we get}$
$\displaystyle 17\lambda(x-1)+2\lambda(y-1)-7\lambda(z+1)=0$
$\displaystyle \text{or,}$
$\displaystyle 17x+2y-7z=26,\text{ which is the required equation of the plane.}$

$\displaystyle \textbf{Question 2:}\quad \text{Find the equation of the plane through the points }(2,1,-1)\text{ and } \\ (-1,3,4)\text{ and perpendicular to the plane }x-2y+4z=10.$
$\displaystyle \text{Also, show that the plane thus obtained contains the line }\overrightarrow{r}=\widehat{i}+3\widehat{j}+4\widehat{k}+\lambda(3\widehat{i}-2\widehat{j}-5\widehat{k}).\quad [\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{The equation of any plane passing through }(2,1,-1)\text{ is}$
$\displaystyle a(x-2)+b(y-1)+c(z+1)=0\qquad \ldots(i)$
$\displaystyle \text{If it passes through }(-1,3,4),\text{ then}$
$\displaystyle a(-1-2)+b(3-1)+c(4+1)=0$
$\displaystyle \Rightarrow -3a+2b+5c=0\qquad \ldots(ii)$
$\displaystyle \text{If plane (i) is perpendicular to the plane }x-2y+4z=10,\text{ then}$
$\displaystyle a-2b+4c=0\qquad \ldots(iii)$
$\displaystyle \text{Solving (ii) and (iii) by the method of cross-multiplication, we obtain}$
$\displaystyle \frac{a}{8+10}=\frac{b}{5+12}=\frac{c}{6-2}$
$\displaystyle \Rightarrow \frac{a}{18}=\frac{b}{17}=\frac{c}{4}=\lambda\ (\text{say})$
$\displaystyle \Rightarrow a=18\lambda,\ b=17\lambda\text{ and }c=4\lambda$
$\displaystyle \text{Putting }a=18\lambda,\ b=17\lambda\text{ and }c=4\lambda\text{ in (i), we obtain}$
$\displaystyle 18\lambda(x-2)+17\lambda(y-1)+4\lambda(z+1)=0$
$\displaystyle \Rightarrow 18x+17y+4z=49\qquad \ldots(iv)$
$\displaystyle \text{This is the required equation of the plane.}$
$\displaystyle \text{The coordinates of any point on the line }\overrightarrow{r}=\widehat{i}+3\widehat{j}+4\widehat{k}+\lambda(3\widehat{i}-2\widehat{j}-5\widehat{k})\text{ are}$
$\displaystyle (3\lambda-1,\,-2\lambda+3,\,-5\lambda+4).$
$\displaystyle \text{Substituting }x=3\lambda-1,\ y=-2\lambda+3,\ z=-5\lambda+4\text{ in (iv), we obtain}$
$\displaystyle \text{LHS}=18(3\lambda-1)+17(-2\lambda+3)+4(-5\lambda+4)=49=\text{RHS}$
$\displaystyle \text{So, }(3\lambda-1,\,-2\lambda+3,\,-5\lambda+4)\text{ lies on plane (iv). Hence, plane in (iv) contains the given line.}$

$\displaystyle \textbf{Question 3:}\quad \text{Find the equation of the plane through the points }(3,4,2)\text{ and } \\ (7,0,6)\text{ and is perpendicular to the plane }2x-5y=15.\quad [\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{The equation of a plane passing through }(3,4,2)\text{ is}$
$\displaystyle a(x-3)+b(y-4)+c(z-2)=0\qquad \ldots(i)$
$\displaystyle \text{This passes through the point }(7,0,6).$
$\displaystyle a(7-3)+b(0-4)+c(6-2)=0$
$\displaystyle \Rightarrow 4a-4b+4c=0\Rightarrow a-b+c=0\qquad \ldots(ii)$
$\displaystyle \text{The plane (i) is perpendicular to the plane }2x-5y+0z=15.$
$\displaystyle 2a+(-5)b+(0)c=0\qquad \ldots(iii)$
$\displaystyle \text{Solving (ii) and (iii) by cross-multiplication, we get}$
$\displaystyle \frac{a}{5}=\frac{b}{2}=\frac{c}{-3}=\lambda\ (\text{say})\Rightarrow a=5\lambda,\ b=2\lambda,\ c=-3\lambda$
$\displaystyle \text{Substituting the values of }a,\ b,\ c\text{ in (i), we get}$
$\displaystyle 5\lambda(x-3)+2\lambda(y-4)-3\lambda(z-2)=0\Rightarrow 5x+2y-3z-17=0$
$\displaystyle \text{This is the equation of the required plane.}$

$\displaystyle \textbf{Question 4:}\quad \text{Find the equation of the plane through the line of intersection of }\overrightarrow{r}\cdot(2\widehat{i}-3\widehat{j}+4\widehat{k})=1 \text{ and }\overrightarrow{r}\cdot(\widehat{i}-\widehat{j})+4=0\text{ and perpendicular to }\overrightarrow{r}\cdot(2\widehat{i}-\widehat{j}+\widehat{k})+8=0.\quad [\text{CBSE 2017}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{The equation of any plane through the line of intersection of the given planes is}$
$\displaystyle [\overrightarrow{r}\cdot(2\widehat{i}-3\widehat{j}+4\widehat{k})-1]+\lambda[\overrightarrow{r}\cdot(\widehat{i}-\widehat{j})+4]=0$
$\displaystyle \text{or, }\overrightarrow{r}\cdot[(2+\lambda)\widehat{i}-(3+\lambda)\widehat{j}+4\widehat{k}]=1-4\lambda\qquad \ldots(i)$
$\displaystyle \text{If plane (i) is perpendicular to }\overrightarrow{r}\cdot(2\widehat{i}-\widehat{j}+\widehat{k})+8=0,\text{ then}$
$\displaystyle [(2+\lambda)\widehat{i}-(3+\lambda)\widehat{j}+4\widehat{k}]\cdot(2\widehat{i}-\widehat{j}+\widehat{k})=0$
$\displaystyle \Rightarrow 2(2+\lambda)+(3+\lambda)+4=0\Rightarrow 3\lambda+11=0\Rightarrow \lambda=-\frac{11}{3}$
$\displaystyle \text{Putting }\lambda=-\frac{11}{3}\text{ in (i), we obtain the equation of the required plane as}$
$\displaystyle \overrightarrow{r}\cdot\left\{\left(2-\frac{11}{3}\right)\widehat{i}-\left(3-\frac{11}{3}\right)\widehat{j}+4\widehat{k}\right\}=1+\frac{44}{3}$
$\displaystyle \text{or, }\overrightarrow{r}\cdot(-5\widehat{i}+2\widehat{j}+12\widehat{k})=47.$

$\displaystyle \textbf{Question 5:}\quad \text{Find the cartesian as well as vector equations of the planes through} \\ \text{the intersection of the} \text{planes }\overrightarrow{r}\cdot(2\widehat{i}+6\widehat{j})+12=0\text{ and }\overrightarrow{r}\cdot(3\widehat{i}-\widehat{j}+4\widehat{k})=0 \\ \text{which } \text{are at a unit distance from the origin.}\quad [\text{CBSE 2005, 2013}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{The equation of the planes through the intersection of the planes }\overrightarrow{r}\cdot(2\widehat{i}+6\widehat{j})+12=0$
$\displaystyle \text{ and }\overrightarrow{r}\cdot(3\widehat{i}-\widehat{j}+4\widehat{k})=0\text{ is}$
$\displaystyle [\overrightarrow{r}\cdot(2\widehat{i}+6\widehat{j})+12]+\lambda[\overrightarrow{r}\cdot(3\widehat{i}-\widehat{j}+4\widehat{k})]=0$
$\displaystyle \text{or, }\overrightarrow{r}\cdot\{(2+3\lambda)\widehat{i}+(6-\lambda)\widehat{j}+4\lambda\widehat{k}\}+12=0.$
$\displaystyle \overrightarrow{r}\cdot\{(-2-3\lambda)\widehat{i}+(\lambda-6)\widehat{j}+(-4\lambda)\widehat{k}\}=12$
$\displaystyle \text{or, }\frac{\overrightarrow{r}\cdot\{(-2-3\lambda)\widehat{i}+(\lambda-6)\widehat{j}+(-4\lambda)\widehat{k}\}}{\sqrt{(2+3\lambda)^{2}+(\lambda-6)^{2}+(4\lambda)^{2}}}=\frac{12}{\sqrt{(2+3\lambda)^{2}+(\lambda-6)^{2}+(4\lambda)^{2}}}$
$\displaystyle \text{This is the normal form of plane (i) and its distance from the origin is}$
$\displaystyle \frac{12}{\sqrt{(2+3\lambda)^{2}+(\lambda-6)^{2}+(4\lambda)^{2}}}$
$\displaystyle \text{It is given that the plane (i) is at a unit distance from the origin.}$
$\displaystyle \therefore \frac{12}{\sqrt{(2+3\lambda)^{2}+(\lambda-6)^{2}+(4\lambda)^{2}}}=1$
$\displaystyle \Rightarrow 144=(2+3\lambda)^{2}+(\lambda-6)^{2}+(4\lambda)^{2}\Rightarrow 26\lambda^{2}=104\Rightarrow \lambda^{2}=4\Rightarrow \lambda=\pm2$
$\displaystyle \text{Putting the values of }\lambda\text{ in (i), we obtain}$
$\displaystyle \overrightarrow{r}\cdot(8\widehat{i}+4\widehat{j}+8\widehat{k})+12=0\text{ and }\overrightarrow{r}\cdot(-4\widehat{i}+8\widehat{j}-8\widehat{k})+12=0$
$\displaystyle \text{as the equations of the required planes.}$
$\displaystyle \text{These equations can also be written as}$
$\displaystyle \overrightarrow{r}\cdot(2\widehat{i}+\widehat{j}+2\widehat{k})+3=0\text{ and }\overrightarrow{r}\cdot(-\widehat{i}+2\widehat{j}-2\widehat{k})+3=0$

$\displaystyle \textbf{Question 6:}\quad \text{Find the distance between the point }P(6,5,9)\text{ and the plane determined} \\ \text{by the points }A(3,-1,2),\ B(5,2,4)\text{ and }C(-1,-1,6).\quad [\text{CBSE 2010, 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{The equation of a plane passing through }A(3,-1,2)\text{ is}$
$\displaystyle a(x-3)+b(y+1)+c(z-2)=0\qquad \ldots(i)$
$\displaystyle \text{If this plane passes through }B(5,2,4)\text{ and }C(-1,-1,6).\text{ Then,}$
$\displaystyle 2a+3b+2c=0$
$\displaystyle \text{and, }-4a+0b+4c=0$
$\displaystyle \text{Using cross-multiplication, we obtain}$
$\displaystyle \frac{a}{12}=\frac{b}{-16}=\frac{c}{12}\text{ or, }\frac{a}{3}=\frac{b}{-4}=\frac{c}{3}=\lambda\ (\text{say})\Rightarrow a=3\lambda,\ b=-4\lambda,\ c=3\lambda$
$\displaystyle \text{Substituting the values of }a,\ b,\ c\text{ in (i), we obtain }3(x-3)-4(y+1)+3(z-2)=0\text{ or,}$
$\displaystyle 3x-4y+3z-19=0\text{ as the equation of the plane passing through }A,\ B\text{ and }C.$
$\displaystyle \text{The distance of }P(6,5,9)\text{ from this plane is given by}$
$\displaystyle p=\frac{|18-20+27-19|}{\sqrt{9+16+9}}=\frac{6}{\sqrt{34}}$

$\displaystyle \textbf{Question 7:}\quad \text{Find the equation of a plane passing through the point } P(6,5,9)\text{ and} \\ \text{parallel to the plane} \text{determined by the points }A(3,-1,2),\ B(5,2,4)\text{ and }C(-1,-1,6). \\ \text{ Also, find the distance of this plane} \text{from the point }A.\quad [\text{CBSE 2015}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{A vector }\overrightarrow{n}\text{ normal to the plane determined by the points }A(3,-1,2),\ B(5,2,4)\text{ and}$
$\displaystyle C(-1,-1,6)\text{ is given by }\overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}.$
$\displaystyle \text{We have,}$
$\displaystyle \overrightarrow{AB}=2\widehat{i}+3\widehat{j}+2\widehat{k}\text{ and }\overrightarrow{AC}=-4\widehat{i}+0\widehat{j}+4\widehat{k}$
$\displaystyle \therefore \overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&3&2\\-4&0&4\end{vmatrix}=12\widehat{i}-16\widehat{j}+12\widehat{k}$
$\displaystyle \text{Clearly, }\overrightarrow{n}=12\widehat{i}-16\widehat{j}+12\widehat{k}\text{ is also normal to the plane passing through }P(6,5,9)\text{ and parallel to}$
$\displaystyle \text{the plane determined by points }A,\ B\text{ and }C.\text{ So, its equation is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n},\text{ where }\overrightarrow{a}=6\widehat{i}+5\widehat{j}+9\widehat{k}$
$\displaystyle \text{or, }\overrightarrow{r}\cdot(12\widehat{i}-16\widehat{j}+12\widehat{k})=(12\widehat{i}-16\widehat{j}+12\widehat{k})\cdot(6\widehat{i}+5\widehat{j}+9\widehat{k})$
$\displaystyle \text{or, }\overrightarrow{r}\cdot(12\widehat{i}-16\widehat{j}+12\widehat{k})=72-80+108$
$\displaystyle \text{or, }\overrightarrow{r}\cdot(3\widehat{i}-4\widehat{j}+3\widehat{k})=25$
$\displaystyle \text{The cartesian equation of this plane is }3x-4y+3z=25.$
$\displaystyle \text{The distance of this plane from the point }A(3,-1,2)\text{ is given by}$
$\displaystyle d=\frac{|3\times3-4\times(-1)+3\times2-25|}{\sqrt{9+16+9}}=\frac{6}{\sqrt{34}}$

$\displaystyle \textbf{Question 8:}\quad \text{Show that the line whose vector equation is } \\ \overrightarrow{r}=(2\widehat{i}-2\widehat{j}+3\widehat{k})+\lambda(\widehat{i}-\widehat{j}+4\widehat{k})\text{ is parallel} \text{ to the plane whose vector equation is } \\ \overrightarrow{r}\cdot(\widehat{i}+5\widehat{j}+\widehat{k})=5.\ \text{Also, find the distance between them.}\quad [\text{CBSE 2001C, 2004}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{The given line passes through the point having position vector }\overrightarrow{a}=2\widehat{i}-2\widehat{j}+3\widehat{k}$
$\displaystyle \text{and is parallel to the vector }\overrightarrow{b}=\widehat{i}-\widehat{j}+4\widehat{k}.\ \text{The given plane is normal to the vector}$
$\displaystyle \overrightarrow{n}=\widehat{i}+5\widehat{j}+\widehat{k}.$
$\displaystyle \text{Now, }\overrightarrow{b}\cdot\overrightarrow{n}=(\widehat{i}-\widehat{j}+4\widehat{k})\cdot(\widehat{i}+5\widehat{j}+\widehat{k})=1-5+4=0$
$\displaystyle \text{So, }\overrightarrow{b}\ \text{perpendicular to }\overrightarrow{n}.\ \text{Hence, the given line is parallel to the given plane.}$
$\displaystyle \text{The distance between the line and the parallel plane is the distance between any} \\ \text{point on the line and the given plane. Since the line passes through the point } \\ \overrightarrow{a}=2\widehat{i}-2\widehat{j}+3\widehat{k}.$
$\displaystyle \text{Therefore, if }d\text{ is the distance between the given line and given plane. Then,}$
$\displaystyle d=\text{Length of perpendicular from }\overrightarrow{a}=2\widehat{i}-2\widehat{j}+3\widehat{k}\text{ to the given plane}$
$\displaystyle \overrightarrow{r}=2\widehat{i}-2\widehat{j}+3\widehat{k}+\lambda(\widehat{i}-\widehat{j}+4\widehat{k})$
$\displaystyle \Rightarrow d=\frac{|(2\widehat{i}-2\widehat{j}+3\widehat{k})\cdot(\widehat{i}+5\widehat{j}+\widehat{k})-5|}{\sqrt{1^{2}+5^{2}+1^{2}}}=\frac{|(2-10+3)-5|}{\sqrt{27}}=\frac{10}{\sqrt{27}}$

$\displaystyle \textbf{Question 9:}\quad \text{Find the equations of the line passing through the point }(3,0,1) \\ \text{ and parallel to the planes }x+2y=0\text{ and }3y-z=0.\quad [\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{Let the direction ratios of the required line be proportional to } \\ a,\ b,\ c.\ \text{As it passes through }(3,0,1).\ \text{So, its equations are}$
$\displaystyle \frac{x-3}{a}=\frac{y-0}{b}=\frac{z-1}{c}\qquad \ldots(i)$
$\displaystyle \text{It is given that the line (i) is parallel to the planes }x+2y+0z=0\text{ and }0x+3y-z=0.$
$\displaystyle \therefore a(1)+b(2)+c(0)=0\text{ and }a(0)+b(3)+c(-1)=0$
$\displaystyle \text{Solving these two equations by cross-multiplication, we obtain}$
$\displaystyle \frac{a}{(2)(-1)-(0)(3)}=\frac{b}{(0)(0)-(1)(-1)}=\frac{c}{(1)(3)-(0)(2)}$
$\displaystyle \Rightarrow \frac{a}{-2}=\frac{b}{1}=\frac{c}{3}=\lambda\ (\text{say})\Rightarrow a=-2\lambda,\ b=\lambda,\ c=3\lambda$
$\displaystyle \text{Substituting the values of }a,\ b,\ c\text{ in (i), we obtain that the equations of the required line are}$
$\displaystyle \frac{x-3}{-2}=\frac{y-0}{1}=\frac{z-1}{3}$

$\displaystyle \textbf{Question 10:}\quad \text{Find the equation of the plane passing through the line of intersection} \\ \text{of the planes} 2x+y-z=3,\ 5x-3y+4z+9=0\text{ and parallel to the line } \\ \frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}.\quad [\text{CBSE 2011, 2015}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{The equation of the plane passing through the line of intersection of the planes}$
$\displaystyle 2x+y-z=3\text{ and }5x-3y+4z+9=0\text{ is}$
$\displaystyle (2x+y-z-3)+\lambda(5x-3y+4z+9)=0$
$\displaystyle \Rightarrow x(2+5\lambda)+y(1-3\lambda)+z(4\lambda-1)+9\lambda-3=0\qquad \ldots(i)$
$\displaystyle \text{The plane in (i) is parallel to the line }\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}.$
$\displaystyle \therefore 2(2+5\lambda)+4(1-3\lambda)+5(4\lambda-1)=0\Rightarrow 18\lambda+3=0\Rightarrow \lambda=-\frac{1}{6}$
$\displaystyle \text{Putting the value of }\lambda\text{ in (i), we obtain}$
$\displaystyle x\left(2-\frac{5}{6}\right)+y\left(1+\frac{3}{6}\right)+z\left(-\frac{4}{6}-1\right)-\frac{9}{6}-3=0$
$\displaystyle \text{or, }7x+9y-10z-27=0,\ \text{which is the required equation of the plane.}$

$\displaystyle \textbf{Question 11:}\quad \text{Find the equation of the plane passing through the intersection of} \\ \text{the planes }\overrightarrow{r}\cdot(\widehat{i}+\widehat{j}+\widehat{k})=1 \text{ and }\overrightarrow{r}\cdot(2\widehat{i}+3\widehat{j}-\widehat{k})+4=0\text{ and parallel to the }x\text{-axis.} \\ \quad [\text{CBSE 2011}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{The equation of a plane passing through the intersection of the planes }\overrightarrow{r}\cdot(\widehat{i}+\widehat{j}+\widehat{k})=1$
$\displaystyle \text{ and }\overrightarrow{r}\cdot(2\widehat{i}+3\widehat{j}-\widehat{k})+4=0\text{ is}$
$\displaystyle \{\overrightarrow{r}\cdot(\widehat{i}+\widehat{j}+\widehat{k})-1\}+\lambda\{\overrightarrow{r}\cdot(2\widehat{i}+3\widehat{j}-\widehat{k})+4\}=0$
$\displaystyle \text{or, }\overrightarrow{r}\cdot\{(2\lambda+1)\widehat{i}+(3\lambda+1)\widehat{j}+(1-\lambda)\widehat{k}\}+(4\lambda-1)=0\qquad \ldots(i)$
$\displaystyle \text{It is given that the plane (i) is parallel to the }x\text{-axis i.e. the vector }\widehat{i}.$
$\displaystyle \therefore \{(2\lambda+1)\widehat{i}+(3\lambda+1)\widehat{j}+(1-\lambda)\widehat{k}\}\cdot\widehat{i}=0\Rightarrow 2\lambda+1=0\Rightarrow \lambda=-\frac{1}{2}$
$\displaystyle \text{Putting }\lambda=-\frac{1}{2}\text{ in (i), we get}$
$\displaystyle \overrightarrow{r}\cdot\left(-\frac{1}{2}\widehat{i}+\frac{3}{2}\widehat{k}\right)-3=0\text{ or, }\overrightarrow{r}\cdot(-\widehat{j}+3\widehat{k})=6,\text{ which is the required equation of the plane.}$

$\displaystyle \textbf{Question 12:}\quad \text{Find the equation of the plane passing through the point }A(1,2,1) \\ \text{and perpendicular to the line joining}\text{the points }P(1,4,2)\text{ and } Q(2,3,5). \text{ Also, find the} \\ \text{distance of this plane from the line } \frac{x+3}{2}=\frac{y-5}{-1}=\frac{z-7}{1}. \quad [\text{CBSE 2010, 2011}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{The direction ratios of }PQ\text{ are proportional to }2-1,\ 3-4,\ 5-2\text{ i.e. }1,-1,3.$
$\displaystyle \text{So, the equation of the plane passing through }A(1,2,1)\text{ and perpendicular to }PQ\text{ is}$
$\displaystyle 1(x-1)+(-1)(y-2)+3(z-1)=0\text{ or, }x-y+3z=2\qquad \ldots(i)$
$\displaystyle \text{The given line is parallel to the vector }\overrightarrow{b}=2\widehat{i}-\widehat{j}-\widehat{k}\text{ and the plane (i) is normal to the vector}$
$\displaystyle \overrightarrow{n}=\widehat{i}-\widehat{j}+3\widehat{k}\text{ such that }\overrightarrow{b}\cdot\overrightarrow{n}=0.\text{ So, given line is parallel to the plane (i).}$
$\displaystyle \text{The distance between the plane (i) and the given line is the distance of any point on the} \\ \text{line from the plane (i).}$
$\displaystyle \text{The line passes through the point }(-3,5,7).$
$\displaystyle \text{So, required distance }=\text{Length of perpendicular from }(-3,5,7)\text{ on plane (i).}$
$\displaystyle =\frac{|-3-5+21-2|}{\sqrt{1^{2}+(-1)^{2}+3^{2}}}=\frac{11}{\sqrt{11}}=\sqrt{11}$

$\displaystyle \textbf{Question 13:}\quad \text{Find the coordinates of the point where the line through the points } \\ A(3,4,1)\text{ and }B(5,1,6)\text{ crosses the }XY\text{-plane.}\quad [\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{The equation of the line passing through }A\text{ and }B$
$\displaystyle \frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1}{6-1}\text{ or, }\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}$
$\displaystyle \text{The coordinates of any point on this line are given by }\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}=\lambda\Rightarrow x=2\lambda+3,\ y=-3\lambda+4,\ z=5\lambda+1$
$\displaystyle \text{So, }(2\lambda+3,\,-3\lambda+4,\ 5\lambda+1)\text{ are coordinates of any point on the line passing through }A\text{ and }B.$
$\displaystyle \text{If it lies on }XY\text{-plane i.e. }z=0,\text{ then }5\lambda+1=0\Rightarrow \lambda=-\frac{1}{5}$
$\displaystyle \text{Thus, the coordinates of required point are }\left(2\times\left(-\frac{1}{5}\right)+3,\,-3\times\left(-\frac{1}{5}\right)+4,\ 5\times\left(-\frac{1}{5}\right)+1\right) \\ \text{ i.e. }\left(\frac{13}{5},\ \frac{23}{5},\ 0\right)$

$\displaystyle \textbf{Question 14:}\quad \text{Find the distance of the point }(1,-2,3)\text{ from the plane } x-y+z=5 \text{ measured parallel to the line whose direction cosines are proportional to } \\ 2,3,-6.\quad [\text{CBSE 2015}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{We have to find the distance }PQ\text{ of point }P(1,-2,3)\text{ from the plane }x-y+z-5=0 \text{measured parallel to the line passing through }A\text{ and }B\text{ whose} \\ \text{direction cosines are proportional to }2,3,-6.$
$\displaystyle \text{Clearly, direction cosines of }PQ\text{ are also proportional to }2,3,-6.\text{ So, equations of }PQ\text{ are given by}$
$\displaystyle \frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}$
$\displaystyle \text{The coordinates of }Q\text{ are given by }\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r$
$\displaystyle \text{or, }(2r+1,\ 3r-2,\ -6r+3)$
$\displaystyle \text{We observe that the point }Q\text{ lies on the plane }x-y+z=5.$
$\displaystyle \therefore 2r+1-3r+2-6r+3=5\Rightarrow -7r=-1\Rightarrow r=\frac{1}{6}$
$\displaystyle \text{Putting }r=\frac{1}{6}\text{ in }(2r+1,\ 3r-2,\ -6r+3),\text{ we obtain that the coordinates of }Q\text{ are }\left(\frac{4}{3},\ -\frac{3}{2},\ 2\right)$
$\displaystyle \text{Hence, required distance }=PQ=\sqrt{\left(\frac{4}{3}-1\right)^{2}+\left(-\frac{3}{2}+2\right)^{2}+(2-3)^{2}}=\sqrt{\frac{1}{9}+\frac{1}{4}+1}=\frac{7}{6}\text{ units.}$

$\displaystyle \textbf{Question 15:}\quad \text{Find the vector equation of the plane that contains the lines }\overrightarrow{r}=(\widehat{i}+\widehat{j})+\lambda(\widehat{i}+2\widehat{j}-\widehat{k})\text{and }\overrightarrow{r}=(\widehat{i}+\widehat{j})+\mu(-\widehat{i}+\widehat{j}-2\widehat{k}).\ \text{Also, find the length of the} \\ \text{perpendicular drawn from the point} (2,1,4)\text{ to the plane thus obtained.}\quad [\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{The two given lines pass through the point having position vector }\overrightarrow{a}=\widehat{i}+\widehat{j}\text{ and are}$
$\displaystyle \text{parallel to the vectors }\overrightarrow{b_{1}}=\widehat{i}+2\widehat{j}-\widehat{k}\text{ and }\overrightarrow{b_{2}}=-\widehat{i}+\widehat{j}-2\widehat{k}\text{ respectively. Therefore, the plane}$
$\displaystyle \text{containing the given lines also passes through the point with position vector }\overrightarrow{a}=\widehat{i}+\widehat{j}.\ \text{Since the plane contains the lines which are parallel to the vectors }\overrightarrow{b_{1}}$
$\displaystyle \text{and }\overrightarrow{b_{2}}\text{ respectively. Therefore, the plane is normal to the vector}$
$\displaystyle \overrightarrow{n}=\overrightarrow{b_{1}}\times\overrightarrow{b_{2}}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&2&-1\\-1&1&-2\end{vmatrix}=-3\widehat{i}+3\widehat{j}+3\widehat{k}$
$\displaystyle \text{Thus, the vector equation of the required plane is}$
$\displaystyle (\overrightarrow{r}-\overrightarrow{a})\cdot\overrightarrow{n}=0\text{ or, }\overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{or, }\overrightarrow{r}\cdot(-3\widehat{i}+3\widehat{j}+3\widehat{k})=(\widehat{i}+\widehat{j})\cdot(-3\widehat{i}+3\widehat{j}+3\widehat{k})$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(-3\widehat{i}+3\widehat{j}+3\widehat{k})=-3+3$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(-\widehat{i}+\widehat{j}+\widehat{k})=0.$
$\displaystyle \text{The length of perpendicular from }P(2,1,4)\text{ to the above plane is given by}$
$\displaystyle d=\frac{|(2\widehat{i}+\widehat{j}+4\widehat{k})\cdot(-\widehat{i}+\widehat{j}+\widehat{k})|}{\sqrt{(-1)^{2}+1^{2}+1^{2}}}=\frac{|-2+1+4|}{\sqrt{3}}=\sqrt{3}$

$\displaystyle \textbf{Question 16:}\quad \text{If this lines }\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\text{ and }\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\text{ intersect, then find the value of }k \text{and hence find the equation of the} \\ \text{plane containing these lines.}\quad [\text{CBSE 2015}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{We know that the lines }\frac{x-x_{1}}{l_{1}}=\frac{y-y_{1}}{m_{1}}=\frac{z-z_{1}}{n_{1}}\text{ and }\frac{x-x_{2}}{l_{2}}=\frac{y-y_{2}}{m_{2}}=\frac{z-z_{2}}{n_{2}}\text{ are}$
$\displaystyle \text{coplanar, if }\begin{vmatrix}x_{2}-x_{1}&y_{2}-y_{1}&z_{2}-z_{1}\\l_{1}&m_{1}&n_{1}\\l_{2}&m_{2}&n_{2}\end{vmatrix}=0$
$\displaystyle \text{and, the equation of the plane containing them is}$
$\displaystyle \begin{vmatrix}x-x_{1}&y-y_{1}&z-z_{1}\\l_{1}&m_{1}&n_{1}\\l_{2}&m_{2}&n_{2}\end{vmatrix}=0\text{ or, }\begin{vmatrix}x-x_{2}&y-y_{2}&z-z_{2}\\l_{1}&m_{1}&n_{1}\\l_{2}&m_{2}&n_{2}\end{vmatrix}=0$
$\displaystyle \text{Here, }x_{1}=1,\ y_{1}=-1,\ z_{1}=1,\ x_{2}=3,\ y_{2}=k,\ z_{2}=0,\ l_{1}=2,\ m_{1}=3,\ n_{1}=4,\ l_{2}=1,\ m_{2}=2,\ n_{2}=1$
$\displaystyle \text{If given lines intersect, then they must be coplanar.}$
$\displaystyle \therefore \begin{vmatrix}3-1&k+1&0-1\\2&3&4\\1&2&1\end{vmatrix}=0$
$\displaystyle \Rightarrow 2(3-8)-(k+1)(2-4)-1(4-3)=0$
$\displaystyle \Rightarrow -10+2k+2-1=0\Rightarrow 2k-9=0\Rightarrow k=\frac{9}{2}$
$\displaystyle \text{The equation of the plane containing the given lines is}$
$\displaystyle \begin{vmatrix}x-1&y+1&z-1\\2&3&4\\1&2&1\end{vmatrix}=0$
$\displaystyle \Rightarrow (x-1)(3-8)-(y+1)(2-4)+(z-1)(4-3)=0$
$\displaystyle \Rightarrow -5x+5+2y+2+z-1=0$
$\displaystyle \Rightarrow 5x-2y-z=6$

$\displaystyle \textbf{Question 17:}\quad \text{Find the length and the foot of the perpendicular from the point } \\ (7,14,5)\text{ to the plane }2x+4y-z=2. \text{Also, find the image of the point }P\text{ in the} \\ \text{plane. } [\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$
$\displaystyle \quad \text{Let }M\text{ be the foot of the perpendicular from }P\text{ on the plane }2x+4y-z=2.$
$\displaystyle \text{Then, }PM\text{ is normal to the plane. So, its direction ratios are proportional to }2,4,-1.$
$\displaystyle \text{Since }PM\text{ passes through }P(7,14,5).\ \text{Therefore, its equation is}$
$\displaystyle \frac{x-7}{2}=\frac{y-14}{4}=\frac{z-5}{-1}=r\ (\text{say})$
$\displaystyle \text{Let the coordinates of }M\text{ be }(2r+7,\ 4r+14,\ -r+5).$
$\displaystyle \text{Since }M\text{ lies on the plane }2x+4y-z=2.\ \text{Therefore,}$
$\displaystyle 2(2r+7)+4(4r+14)-(-r+5)=2\Rightarrow 21r+63=0\Rightarrow r=-3$
$\displaystyle \text{So, the coordinates of }M\text{ are }(1,2,8).$
$\displaystyle \therefore PM=\sqrt{(7-1)^{2}+(14-2)^{2}+(5-8)^{2}}=3\sqrt{21}$
$\displaystyle \text{Hence, length of the perpendicular from }P=3\sqrt{21}.$
$\displaystyle \text{Let }Q(x_{1},y_{1},z_{1})\text{ be the image of }P\text{ in the given plane. Then, the coordinates of }M\text{ are}$
$\displaystyle \left(\frac{x_{1}+7}{2},\ \frac{y_{1}+14}{2},\ \frac{z_{1}+5}{2}\right).$
$\displaystyle \text{But, the coordinates of }M\text{ are given as }(1,2,8).$
$\displaystyle \therefore \frac{x_{1}+7}{2}=1,\ \frac{y_{1}+14}{2}=2,\ \frac{z_{1}+5}{2}=8\Rightarrow x_{1}=-5,\ y_{1}=-10,\ z_{1}=11$
$\displaystyle \text{Hence, coordinates of }Q\text{ are }(-5,-10,11).$