\displaystyle \textbf{Question 1: }~\text{Find the image of the point }(0,0,0)\text{ in the plane } \\ 3x+4y-6z+1=0.
\displaystyle \text{Answer:}
\displaystyle \text{Let }Q\text{ be the image of the point }P(0,0,0)\text{ in the plane }3x+4y-6z+1=0.
\displaystyle \text{Then }PQ\text{ is normal to the plane. So, the direction ratios of }PQ\text{ are proportional to }3,4,-6.
\displaystyle \text{Since }PQ\text{ passes through }P(0,0,0)\text{ and has direction ratios proportional to }3,4,-6,\text{ equation of }PQ\text{ is}
\displaystyle \frac{x-0}{3}=\frac{y-0}{4}=\frac{z-0}{-6}=r\ (\text{say}).
\displaystyle \text{Let the coordinates of }Q\text{ be }(3r,4r,-6r).\text{ Let }R\text{ be the mid-point of }PQ.\text{ Then,}
\displaystyle R=\left(\frac{0+3r}{2},\frac{0+4r}{2},\frac{0-6r}{2}\right)=\left(\frac{3r}{2},2r,-3r\right).
\displaystyle \text{Since }R\text{ lies in the plane }3x+4y-6z+1=0,
\displaystyle 3\left(\frac{3r}{2}\right)+4(2r)-6(-3r)+1=0.
\displaystyle \Rightarrow \frac{61r}{2}+1=0.
\displaystyle \Rightarrow r=-\frac{2}{61}.
\displaystyle \text{Substituting this in the coordinates of }Q,\text{ we get}
\displaystyle Q=(3r,4r,-6r)=\left(3\left(-\frac{2}{61}\right),4\left(-\frac{2}{61}\right),-6\left(-\frac{2}{61}\right)\right)=\left(-\frac{6}{61},-\frac{8}{61},\frac{12}{61}\right).

\displaystyle \textbf{Question 2: }~\text{Find the reflection of the point }(1,2,-1)\text{ in the plane } \\ 3x-5y+4z=5.
\displaystyle \text{Answer:}
\displaystyle \text{Let }Q\text{ be the image of the point }P(1,2,-1)\text{ in the plane }3x-5y+4z=5.
\displaystyle \text{Then }PQ\text{ is normal to the plane. So, the direction ratios of }PQ\text{ are proportional to }3,-5,4.
\displaystyle \text{Since }PQ\text{ passes through }P(1,2,-1)\text{ and has direction ratios proportional to }3,-5,4,\text{ equation of }PQ\text{ is}
\displaystyle \frac{x-1}{3}=\frac{y-2}{-5}=\frac{z+1}{4}=r\ (\text{say}).
\displaystyle \text{Let the coordinates of }Q\text{ be }(3r+1,-5r+2,4r-1).\text{ Let }R\text{ be the mid-point of }PQ.\text{ Then,}
\displaystyle R=\left(\frac{3r+1+1}{2},\frac{-5r+2+2}{2},\frac{4r-1-1}{2}\right)=\left(\frac{3r+2}{2},\frac{-5r+4}{2},\frac{4r-2}{2}\right).
\displaystyle \text{Since }R\text{ lies in the plane }3x-5y+4z=5,
\displaystyle 3\left(\frac{3r+2}{2}\right)-5\left(\frac{-5r+4}{2}\right)+4\left(\frac{4r-2}{2}\right)=5.
\displaystyle \Rightarrow 9r+6+25r-20+16r-8=10.
\displaystyle \Rightarrow 50r-22=10.
\displaystyle \Rightarrow 50r=32.
\displaystyle \Rightarrow r=\frac{32}{50}=\frac{16}{25}.
\displaystyle \text{Substituting the value of }r\text{ in the coordinates of }Q,\text{ we get}
\displaystyle Q=(3r+1,-5r+2,4r-1)=\left(3\left(\frac{16}{25}\right)+1,-5\left(\frac{16}{25}\right)+2,4\left(\frac{16}{25}\right)-1\right)=\left(\frac{73}{25},-\frac{6}{5},\frac{39}{25}\right).

\displaystyle \textbf{Question 3: }~\text{Find the coordinates of the foot of the perpendicular drawn} \\ \text{from the point }(5,4,2) \text{ to the line }\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}. \text{ Hence or otherwise} \\ \text{deduce the lengthof the perpendicular.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }M\text{ be the foot of the perpendicular of the point }P(5,4,2)\text{ on the line }\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}.
\displaystyle \text{Therefore, its equation is}
\displaystyle \frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}=r.
\displaystyle \text{Then, }M\text{ is in the form }(2r-1,3r+3,-r+1).
\displaystyle \text{Direction ratios of }MP\text{ are }2r-1-5,3r+3-4,-r+1-2\text{ or }2r-6,3r-1,-r-1.
\displaystyle \text{Since }MP\text{ is perpendicular to the given line }(2,3,-1),
\displaystyle 2(2r-6)+3(3r-1)-1(-r-1)=0.
\displaystyle \text{(Because }a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0\text{)}
\displaystyle \Rightarrow 4r-12+9r-3+r+1=0.
\displaystyle \Rightarrow 14r-14=0.
\displaystyle \Rightarrow r=1.
\displaystyle \text{So, }M=(2r-1,3r+3,-r+1)=(2(1)-1,3(1)+3,-1+1)=(1,6,0).
\displaystyle \text{Length of the perpendicular }MP=\sqrt{(1-5)^{2}+(6-4)^{2}+(0-2)^{2}}=\sqrt{16+4+4}=\sqrt{24}=2\sqrt{6}\text{ units.}

\displaystyle \textbf{Question 4: }~\text{Find the image of the point with position vector }3\widehat{i}+\widehat{j}+2\widehat{k}\text{ in the plane }\overrightarrow{r}\cdot(2\widehat{i}-\widehat{j}+\widehat{k})=4. \text{ Also, find the position vectors of the foot of the perpendicular and the} \\ \text{equation of the perpendicular line through }3\widehat{i}+\widehat{j}+2\widehat{k}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }Q\text{ be the image of the point }P(3\widehat{i}+\widehat{j}+2\widehat{k})\text{ in the plane }\overrightarrow{r}\cdot(2\widehat{i}-\widehat{j}+\widehat{k})=4.
\displaystyle \text{Since }PQ\text{ passes through }P\text{ and is normal to the given plane, it is parallel to the normal vector }2\widehat{i}-\widehat{j}+\widehat{k}.\text{ So, the equation of }PQ\text{ is}
\displaystyle \overrightarrow{r}=(3\widehat{i}+\widehat{j}+2\widehat{k})+\lambda(2\widehat{i}-\widehat{j}+\widehat{k}).
\displaystyle \text{As }Q\text{ lies on }PQ,\text{ let the position vector of }Q\text{ be }(3+2\lambda)\widehat{i}+(1-\lambda)\widehat{j}+(2+\lambda)\widehat{k}.
\displaystyle \text{Let }R\text{ be the mid-point of }PQ.\text{ Then, the position vector of }R\text{ is}
\displaystyle \frac{(3+2\lambda)\widehat{i}+(1-\lambda)\widehat{j}+(2+\lambda)\widehat{k}+(3\widehat{i}+\widehat{j}+2\widehat{k})}{2}.
\displaystyle =\frac{(6+2\lambda)\widehat{i}+(2-\lambda)\widehat{j}+(4+\lambda)\widehat{k}}{2}.
\displaystyle =(3+\lambda)\widehat{i}+\left(1-\frac{\lambda}{2}\right)\widehat{j}+\left(2+\frac{\lambda}{2}\right)\widehat{k}.
\displaystyle \text{Since }R\text{ lies in the plane }\overrightarrow{r}\cdot(2\widehat{i}-\widehat{j}+\widehat{k})=4,
\displaystyle \left[(3+\lambda)\widehat{i}+\left(1-\frac{\lambda}{2}\right)\widehat{j}+\left(2+\frac{\lambda}{2}\right)\widehat{k}\right]\cdot(2\widehat{i}-\widehat{j}+\widehat{k})=4.
\displaystyle \Rightarrow 6+2\lambda-1+\frac{\lambda}{2}+2+\frac{\lambda}{2}=4.
\displaystyle \Rightarrow 7+3\lambda=4.
\displaystyle \Rightarrow 3\lambda=-3.
\displaystyle \Rightarrow \lambda=-1.
\displaystyle \text{Putting }\lambda=-1\text{ in }Q,\text{ we get}
\displaystyle Q=(3+2(-1))\widehat{i}+(1-(-1))\widehat{j}+(2+(-1))\widehat{k}=\widehat{i}+2\widehat{j}+\widehat{k}\text{ or }(1,2,1).
\displaystyle \text{Therefore, by putting }\lambda=-1\text{ in }R,\text{ we get}
\displaystyle R=(3-1)\widehat{i}+\left(1-\frac{-1}{2}\right)\widehat{j}+\left(2+\frac{-1}{2}\right)\widehat{k}.
\displaystyle =2\widehat{i}+\frac{3}{2}\widehat{j}+\frac{3}{2}\widehat{k}.

\displaystyle \textbf{Question 5: }~\text{Find the coordinates of the foot of the perpendicular from the point } \\ (1,1,2)\text{ to the plane }2x-2y+4z+5=0. \text{ Also, find the length of the perpendicular.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }M\text{ be the foot of the perpendicular of the point }P(1,1,2)\text{ in the plane }2x-2y+4z+5=0.
\displaystyle \text{Then }PM\text{ is normal to the plane. So, the direction ratios of }PM\text{ are proportional to }2,-2,4.
\displaystyle \text{Since }PM\text{ passes through }P(1,1,2)\text{ and has direction ratios proportional to }2,-2,4,\text{ equation of }PM\text{ is}
\displaystyle \frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{4}=r\ (\text{say}).
\displaystyle \text{Let the coordinates of }M\text{ be }(2r+1,-2r+1,4r+2).
\displaystyle \text{Since }M\text{ lies in the plane }2x-2y+4z+5=0,
\displaystyle 2(2r+1)-2(-2r+1)+4(4r+2)+5=0.
\displaystyle \Rightarrow 4r+2+4r-2+16r+8+5=0.
\displaystyle \Rightarrow 24r+13=0.
\displaystyle \Rightarrow r=-\frac{13}{24}.
\displaystyle \text{Substituting this in the coordinates of }M,\text{ we get}
\displaystyle M=(2r+1,-2r+1,4r+2)=\left(2\left(-\frac{13}{24}\right)+1,-2\left(-\frac{13}{24}\right)+1,4\left(-\frac{13}{24}\right)+2\right)=\left(-\frac{1}{12},\frac{25}{12},-\frac{1}{6}\right).
\displaystyle \text{Now, the length of the perpendicular from }P\text{ onto the given plane}
\displaystyle =\frac{|2(1)-2(1)+4(2)+5|}{\sqrt{4+4+16}}.
\displaystyle =\frac{13}{\sqrt{24}}\text{ units.}

\displaystyle \textbf{Question 6: }~\text{Find the distance of the point }(1,-2,3)\text{ from the plane } \\ x-y+z=5\text{ measured along a line parallel to }\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}. \text{[CBSE\ 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{The given plane is }x-y+z=5.\text{ We need to find the distance of the point from the point }(1,-2,3)\text{ measured along a parallel line}
\displaystyle \frac{x}{2}=\frac{y}{3}=\frac{z}{-6}.
\displaystyle \text{Let the line from the point be }P(1,-2,3)\text{ and meet the plane at point }Q.
\displaystyle \text{Direction ratios of the line from the point }(1,-2,3)\text{ to the given plane will be the same as the given line}
\displaystyle \frac{x}{2}=\frac{y}{3}=\frac{z}{-6}.
\displaystyle \text{So the equation of the line passing through }P\text{ and with the same direction ratios will be}
\displaystyle \frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=\lambda.
\displaystyle \Rightarrow x=2\lambda+1,\ y=3\lambda-2,\ z=-6\lambda+3.
\displaystyle \text{Coordinates of any point on the line }PQ\text{ are}
\displaystyle x=2\lambda+1,\ y=3\lambda-2,\ z=-6\lambda+3.
\displaystyle \text{Now, since }Q\text{ lies on the plane it must satisfy the equation of the plane,}
\displaystyle x-y+z=5.
\displaystyle \text{Therefore, }2\lambda+1-(3\lambda-2)+(-6\lambda+3)=5.
\displaystyle \Rightarrow -7\lambda+6=5.
\displaystyle \Rightarrow \lambda=\frac{1}{7}.
\displaystyle \text{Coordinates of }Q\text{ are}
\displaystyle \left(\frac{2}{7}+1,\frac{3}{7}-2,-\frac{6}{7}+3\right)=\left(\frac{9}{7},-\frac{11}{7},\frac{15}{7}\right).
\displaystyle \text{Using the distance formula we have the length of }PQ\text{ as}
\displaystyle PQ=\sqrt{\left(\frac{9}{7}-1\right)^{2}+\left(-\frac{11}{7}+2\right)^{2}+\left(\frac{15}{7}-3\right)^{2}}.
\displaystyle =\sqrt{\left(\frac{2}{7}\right)^{2}+\left(\frac{3}{7}\right)^{2}+\left(-\frac{6}{7}\right)^{2}}.
\displaystyle =\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}.
\displaystyle =\sqrt{\frac{49}{49}}=1.
\displaystyle \text{Hence }PQ=1.
\displaystyle \text{So, the distance of the point }(1,-2,3)\text{ from the plane }x-y+z=5\text{ is }1.

\displaystyle \textbf{Question 7: }~\text{Find the coordinates of the foot of the perpendicular from the point } \\ (2,3,7)\text{ to the plane }3x-y-z=7. \text{ Also, find the length of the perpendicular.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }M\text{ be the foot of the perpendicular of the point }P(2,3,7)\text{ in the plane }3x-y-z=7.
\displaystyle \text{Then }PM\text{ is normal to the plane. So, the direction ratios of }PM\text{ are proportional to }3,-1,-1.
\displaystyle \text{Since }PM\text{ passes through }P(2,3,7)\text{ and has direction ratios proportional to }3,-1,-1,\text{ equation of }PM\text{ is}
\displaystyle \frac{x-2}{3}=\frac{y-3}{-1}=\frac{z-7}{-1}=r\ (\text{say}).
\displaystyle \text{Let the coordinates of }M\text{ be }(3r+2,-r+3,-r+7).
\displaystyle \text{Since }M\text{ lies in the plane }3x-y-z=7,
\displaystyle 3(3r+2)-(-r+3)-(-r+7)=7.
\displaystyle \Rightarrow 9r+6+r-3+r-7=7.
\displaystyle \Rightarrow 11r-4=7.
\displaystyle \Rightarrow 11r=11.
\displaystyle \Rightarrow r=1.
\displaystyle \text{Substituting this in the coordinates of }M,\text{ we get}
\displaystyle M=(3r+2,-r+3,-r+7)=(3(1)+2,-1+3,-1+7)=(5,2,6).
\displaystyle \text{Now, the length of the perpendicular from }P\text{ onto the given plane}
\displaystyle =\frac{|3(2)-3-7-7|}{\sqrt{9+1+1}}.
\displaystyle =\frac{11}{\sqrt{11}}.
\displaystyle =\sqrt{11}\text{ units.}

\displaystyle \textbf{Question 8: }~\text{Find the image of the point }(1,3,4)\text{ in the plane } \\ 2x-y+z+3=0.
\displaystyle \text{Answer:}
\displaystyle \text{Let }Q\text{ be the image of the point }P(1,3,4)\text{ in the plane }2x-y+z+3=0.
\displaystyle \text{Then }PQ\text{ is normal to the plane. So, the direction ratios of }PQ\text{ are proportional to }2,-1,1.
\displaystyle \text{Since }PQ\text{ passes through }P(1,3,4)\text{ and has direction ratios proportional to }2,-1,1,\text{ equation of }PQ\text{ is}
\displaystyle \frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=r\ (\text{say}).
\displaystyle \text{Let the coordinates of }Q\text{ be }(2r+1,-r+3,r+4).\text{ Let }R\text{ be the mid-point of }PQ.\text{ Then,}
\displaystyle R=\left(\frac{2r+1+1}{2},\frac{-r+3+3}{2},\frac{r+4+4}{2}\right)=\left(r+1,\frac{-r+6}{2},\frac{r+8}{2}\right).
\displaystyle \text{Since }R\text{ lies in the plane }2x-y+z+3=0,
\displaystyle 2(r+1)-\left(\frac{-r+6}{2}\right)+\left(\frac{r+8}{2}\right)+3=0.
\displaystyle \Rightarrow 4r+4+r-6+r+8+6=0.
\displaystyle \Rightarrow 6r+12=0.
\displaystyle \Rightarrow r=-2.
\displaystyle \text{Substituting this in the coordinates of }Q,\text{ we get}
\displaystyle Q=(2r+1,-r+3,r+4)=(2(-2)+1,-(-2)+3,-2+4)=(-3,5,2).

\displaystyle \textbf{Question 9: }~\text{Find the distance of the point with position vector }-\widehat{i}-5\widehat{j}-10\widehat{k}\text{ from the point of intersection of the line }\overrightarrow{r}=(2\widehat{i}-\widehat{j}+2\widehat{k})+\lambda(3\widehat{i}+4\widehat{j}+12\widehat{k})\text{ with the plane }\overrightarrow{r}\cdot(\widehat{i}-\widehat{j}+\widehat{k})=5. \text{[CBSE\ 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{The given equation of the line is}
\displaystyle \overrightarrow{r}=(2\widehat{i}-\widehat{j}+2\widehat{k})+\lambda(3\widehat{i}+4\widehat{j}+2\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}=(2+3\lambda)\widehat{i}+(-1+4\lambda)\widehat{j}+(2+2\lambda)\widehat{k}.
\displaystyle \text{The coordinates of any point on this line are of the form }(2+3\lambda,-1+4\lambda,2+2\lambda).
\displaystyle \text{Since this point lies on the plane }\overrightarrow{r}\cdot(\widehat{i}-\widehat{j}+\widehat{k})=5,
\displaystyle \left[(2+3\lambda)\widehat{i}+(-1+4\lambda)\widehat{j}+(2+2\lambda)\widehat{k}\right]\cdot(\widehat{i}-\widehat{j}+\widehat{k})=5.
\displaystyle \Rightarrow 2+3\lambda+1-4\lambda+2+2\lambda=5.
\displaystyle \Rightarrow 5+\lambda=5.
\displaystyle \Rightarrow \lambda=0.
\displaystyle \text{So, the coordinates of the point are}
\displaystyle (2+3\lambda,-1+4\lambda,2+2\lambda).
\displaystyle =(2+0,-1+0,2+0).
\displaystyle =(2,-1,2).
\displaystyle \text{The coordinates of the point corresponding to the position vector }-\widehat{i}-5\widehat{j}-10\widehat{k}\text{ are }(-1,-5,-10).
\displaystyle \text{Distance between }(2,-1,2)\text{ and }(-1,-5,-10)
\displaystyle =\sqrt{(-1-2)^{2}+(-5+1)^{2}+(-10-2)^{2}}.
\displaystyle =\sqrt{9+16+144}.
\displaystyle =13\text{ units.}

\displaystyle \textbf{Question 10: }~\text{Find the length and the foot of the perpendicular from the point }(1,1,2) \\ \text{ to the plane }\overrightarrow{r}\cdot(\widehat{i}-2\widehat{j}+4\widehat{k})+5=0. \text{[CBSE\ 2002C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }M\text{ be the foot of the perpendicular of the point }P(1,1,2)\text{ in the plane }\overrightarrow{r}\cdot(\widehat{i}-2\widehat{j}+4\widehat{k})+5=0\text{ or }x-2y+4z+5=0.
\displaystyle \text{Then }PM\text{ is the normal to the plane. So, the direction ratios of }PM\text{ are proportional to }1,-2,4.
\displaystyle \text{Since }PM\text{ passes through }P(1,1,2)\text{ and has direction ratios proportional to }1,-2,4,\text{ equation of }PM\text{ is}
\displaystyle \frac{x-1}{1}=\frac{y-1}{-2}=\frac{z-2}{4}=r\ (\text{say}).
\displaystyle \text{Let the coordinates of }M\text{ be }(r+1,-2r+1,4r+2).
\displaystyle \text{Since }M\text{ lies in the plane }x-2y+4z+5=0,
\displaystyle (r+1)-2(-2r+1)+4(4r+2)+5=0.
\displaystyle \Rightarrow r+1+4r-2+16r+8+5=0.
\displaystyle \Rightarrow 21r+12=0.
\displaystyle \Rightarrow r=-\frac{12}{21}=-\frac{4}{7}.
\displaystyle \text{Substituting this in the coordinates of }M,\text{ we get}
\displaystyle M=(r+1,-2r+1,4r+2)=\left(-\frac{4}{7}+1,-2\left(-\frac{4}{7}\right)+1,4\left(-\frac{4}{7}\right)+2\right)=\left(\frac{3}{7},\frac{15}{7},-\frac{2}{7}\right).
\displaystyle \text{Now, the length of the perpendicular from }P\text{ onto the given plane}
\displaystyle =\frac{|1(1)-2(1)+4(2)+5|}{\sqrt{1+4+16}}.
\displaystyle =\frac{12}{\sqrt{21}}\text{ units.}

\displaystyle \textbf{Question 11: }~\text{Find the coordinates of the foot of the perpendicular and the} \\ \text{perpendicular distance of the point }P(3,2,1)\text{ from the plane }2x-y+z+1=0. \\ \text{ Find also the image of the point in the plane. \ [CBSE\ 2010,\ 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }M\text{ be the foot of the perpendicular of the point }P(3,2,1)\text{ in the plane }2x-y+z+1=0.
\displaystyle \text{Then }PM\text{ is the normal to the plane. So, the direction ratios of }PM\text{ are proportional to }2,-1,1.
\displaystyle \text{Since }PM\text{ passes through }P(3,2,1)\text{ and has direction ratios proportional to }2,-1,1,\text{ equation of }PM\text{ is}
\displaystyle \frac{x-3}{2}=\frac{y-2}{-1}=\frac{z-1}{1}=r\ (\text{say}).
\displaystyle \text{Let the coordinates of }M\text{ be }(2r+3,-r+2,r+1).
\displaystyle \text{Since }M\text{ lies in the plane }2x-y+z+1=0,
\displaystyle 2(2r+3)-(-r+2)+(r+1)+1=0.
\displaystyle \Rightarrow 4r+6+r-2+r+2=0.
\displaystyle \Rightarrow 6r+6=0.
\displaystyle \Rightarrow r=-1.
\displaystyle \text{Substituting the value of }r\text{ in the coordinates of }M,\text{ we get}
\displaystyle M=(2r+3,-r+2,r+1)=(2(-1)+3,-(-1)+2,-1+1)=(1,3,0).
\displaystyle \text{Now, the length of the perpendicular from }P\text{ onto the given plane}
\displaystyle =\frac{|2(3)-2(1)+1+1|}{\sqrt{4+1+1}}.
\displaystyle =\frac{6}{\sqrt{6}}.
\displaystyle =\sqrt{6}\text{ units.}

\displaystyle \textbf{Question 12: }~\text{Find the direction cosines of the unit vector perpendicular to the plane } \\ \overrightarrow{r}\cdot(6\widehat{i}-3\widehat{j}-2\widehat{k})+1=0\text{ passing through the origin.  }
\displaystyle \text{Answer:}
\displaystyle \text{For the unit vector perpendicular to the given plane, we need to convert the given equation of plane into normal form.}
\displaystyle \text{The given equation of the plane is}
\displaystyle \overrightarrow{r}\cdot(6\widehat{i}-3\widehat{j}-2\widehat{k})+1=0.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(6\widehat{i}-3\widehat{j}-2\widehat{k})=-1.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(-6\widehat{i}+3\widehat{j}+2\widehat{k})=1\ \ldots(1).
\displaystyle \text{Now, }\sqrt{(-6)^{2}+3^{2}+2^{2}}=\sqrt{36+9+4}=7.
\displaystyle \text{Dividing (1) by }7,\text{ we get}
\displaystyle \overrightarrow{r}\cdot\left(-\frac{6}{7}\widehat{i}+\frac{3}{7}\widehat{j}+\frac{2}{7}\widehat{k}\right)=\frac{1}{7},\ \text{which is in the normal form }\overrightarrow{r}\cdot\widehat{n}=d.
\displaystyle \text{where the unit vector normal to the given plane is }\widehat{n}=-\frac{6}{7}\widehat{i}+\frac{3}{7}\widehat{j}+\frac{2}{7}\widehat{k}.
\displaystyle \text{So, its direction cosines are }-\frac{6}{7},\frac{3}{7},\frac{2}{7}.

\displaystyle \textbf{Question 13: }~\text{Find the coordinates of the foot of the perpendicular drawn from the} \\ \text{origin to the plane }2x-3y+4z-6=0. 
\displaystyle \text{Answer:}
\displaystyle \text{Let }M\text{ be the foot of the perpendicular of the origin }P(0,0,0)\text{ in the plane }2x-3y+4z-6=0.
\displaystyle \text{Then }PM\text{ is normal to the plane. So, the direction ratios of }PM\text{ are proportional to }2,-3,4.
\displaystyle \text{Since }PM\text{ passes through }P(0,0,0)\text{ and has direction ratios proportional to }2,-3,4,\text{ equation of }PM\text{ is}
\displaystyle \frac{x-0}{2}=\frac{y-0}{-3}=\frac{z-0}{4}=r\ (\text{say}).
\displaystyle \text{Let the coordinates of }M\text{ be }(2r,-3r,4r).
\displaystyle \text{Since }M\text{ lies in the plane }2x-3y+4z-6=0,
\displaystyle 2(2r)-3(-3r)+4(4r)-6=0.
\displaystyle \Rightarrow 4r+9r+16r-6=0.
\displaystyle \Rightarrow 29r-6=0.
\displaystyle \Rightarrow r=\frac{6}{29}.
\displaystyle \text{Substituting the value of }r\text{ in the coordinates of }M,\text{ we get}
\displaystyle M=(2r,-3r,4r)=\left(2\left(\frac{6}{29}\right),-3\left(\frac{6}{29}\right),4\left(\frac{6}{29}\right)\right)=\left(\frac{12}{29},-\frac{18}{29},\frac{24}{29}\right).

\displaystyle \textbf{Question 14: }~\text{Find the length and the foot of perpendicular from the point } \\ (1,\frac{3}{2},2)\text{ to the plane }2x-2y+4z+5=0.   
\displaystyle \text{Answer:}
\displaystyle \text{Let }M\text{ be the foot of the perpendicular from }P\left(1,\frac{3}{2},2\right)\text{ on the plane }2x-2y+4z+5=0.\text{ Then, }PM\text{ is the normal to the plane. So, its direction ratios are proportional to }2,-2,4.
\displaystyle \text{Since }PM\text{ passes through }P\left(1,\frac{3}{2},2\right),\text{ therefore, its equation is}
\displaystyle \frac{x-1}{2}=\frac{y-\frac{3}{2}}{-2}=\frac{z-2}{4}=\lambda\ (\text{say}).
\displaystyle \text{Let the coordinates of }M\text{ be }\left(2\lambda+1,-2\lambda+\frac{3}{2},4\lambda+2\right).
\displaystyle \text{Now, }M\text{ lies in the plane }2x-2y+4z+5=0:
\displaystyle 2(2\lambda+1)-2\left(-2\lambda+\frac{3}{2}\right)+4(4\lambda+2)+5=0.
\displaystyle \Rightarrow 4\lambda+2+4\lambda-3+16\lambda+8+5=0.
\displaystyle \Rightarrow 24\lambda+12=0.
\displaystyle \Rightarrow \lambda=-\frac{1}{2}.
\displaystyle \text{So, the coordinates of }M\text{ are}
\displaystyle \left(2\left(-\frac{1}{2}\right)+1,-2\left(-\frac{1}{2}\right)+\frac{3}{2},4\left(-\frac{1}{2}\right)+2\right)=\left(0,\frac{5}{2},0\right).
\displaystyle \text{Thus, the coordinates of the foot of the perpendicular are }\left(0,\frac{5}{2},0\right).
\displaystyle \text{Now,}
\displaystyle PM=\sqrt{(1-0)^{2}+\left(\frac{3}{2}-\frac{5}{2}\right)^{2}+(2-0)^{2}}=\sqrt{1+1+4}=\sqrt{6}.
\displaystyle \text{Thus, the length of the perpendicular from the given point to the plane is }\sqrt{6}\text{ units.}

\displaystyle \textbf{Question 15: }~\text{Find the position vector of the foot of the perpendicular and the} \\ \text{perpendicular distance from the point }P\text{ with position vector }2\widehat{i}+3\widehat{j}+4\widehat{k} \\ \text{ to the plane }\overrightarrow{r}\cdot(2\widehat{i}+\widehat{j}+3\widehat{k})-26=0. \text{ Also, find the image of }P\text{ in the plane. \ [CBSE\ 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }M\text{ be the foot of the perpendicular drawn from the point }P(2,3,4)\text{ in the plane }\overrightarrow{r}\cdot(2\widehat{i}+\widehat{j}+3\widehat{k})-26=0\text{ or }2x+y+3z-26=0.
\displaystyle \text{Then }PM\text{ is the normal to the plane. So, the direction ratios of }PM\text{ are proportional to }2,1,3.
\displaystyle \text{Since }PM\text{ passes through }P(2,3,4)\text{ and has direction ratios proportional to }2,1,3,\text{ the equation of }PM\text{ is}
\displaystyle \frac{x-2}{2}=\frac{y-3}{1}=\frac{z-4}{3}=r\ (\text{say}).
\displaystyle \text{Let the coordinates of }M\text{ be }(2r+2,r+3,3r+4).
\displaystyle \text{Since }M\text{ lies in the plane }2x+y+3z-26=0,
\displaystyle 2(2r+2)+r+3+3(3r+4)-26=0.
\displaystyle \Rightarrow 4r+4+r+3+9r+12-26=0.
\displaystyle \Rightarrow 14r-7=0.
\displaystyle \Rightarrow r=\frac{1}{2}.
\displaystyle \text{Therefore, the coordinates of }M\text{ are}
\displaystyle (2r+2,r+3,3r+4)=\left(2\left(\frac{1}{2}\right)+2,\frac{1}{2}+3,3\left(\frac{1}{2}\right)+4\right)=\left(3,\frac{7}{2},\frac{11}{2}\right).
\displaystyle \text{Thus, the position vector of the foot of the perpendicular is }3\widehat{i}+\frac{7}{2}\widehat{j}+\frac{11}{2}\widehat{k}.
\displaystyle \text{Now,}
\displaystyle \text{Length of the perpendicular from }P\text{ on to the given plane}
\displaystyle =\frac{|2(2)+1(3)+3(4)-26|}{\sqrt{4+1+9}}.
\displaystyle =\frac{7}{\sqrt{14}}.
\displaystyle =\sqrt{\frac{7}{2}}\text{ units.}
\displaystyle \text{Let }Q(x_{1},y_{1},z_{1})\text{ be the image of point }P\text{ in the given plane.}
\displaystyle \text{Then, the coordinates of }M\text{ are }\left(\frac{x_{1}+2}{2},\frac{y_{1}+3}{2},\frac{z_{1}+4}{2}\right).\text{ But the coordinates of }M\text{ are }\left(3,\frac{7}{2},\frac{11}{2}\right).
\displaystyle \therefore \left(\frac{x_{1}+2}{2},\frac{y_{1}+3}{2},\frac{z_{1}+4}{2}\right)=\left(3,\frac{7}{2},\frac{11}{2}\right).
\displaystyle \Rightarrow \frac{x_{1}+2}{2}=3,\ \frac{y_{1}+3}{2}=\frac{7}{2},\ \frac{z_{1}+4}{2}=\frac{11}{2}.
\displaystyle \Rightarrow x_{1}=4,\ y_{1}=4,\ z_{1}=7.
\displaystyle \text{Thus, the coordinates of the image of the point }P\text{ in the given plane are }(4,4,7).

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