\displaystyle \textbf{Question 1: }~\text{Find the distance of the point }2\widehat{i}-\widehat{j}-4\widehat{k}\text{ from the plane }\overrightarrow{r}\cdot(3\widehat{i}-4\widehat{j}+12\widehat{k})-9=0.
\displaystyle \text{Answer:}
\displaystyle \text{We know that the perpendicular distance }p\text{ of a point of position vector }\vec a\text{ from the plane }\vec r\cdot\vec n=d\text{ is given by}
\displaystyle p=\frac{\left|\vec a\cdot\vec n-d\right|}{\left|\vec n\right|}.
\displaystyle \text{Here, }\vec a=2\hat i-\hat j-4\hat k,\ \vec n=3\hat i-4\hat j+12\hat k,\ d=9.
\displaystyle \text{So, the required distance }p
\displaystyle =\frac{\left|(2\hat i-\hat j-4\hat k)\cdot(3\hat i-4\hat j+12\hat k)-9\right|}{\left|3\hat i-4\hat j+12\hat k\right|}.
\displaystyle =\frac{\left|6+4-48-9\right|}{\sqrt{9+16+144}}.
\displaystyle =\frac{47}{\sqrt{169}}.
\displaystyle =\frac{47}{13} \text{units.}

\displaystyle \textbf{Question 2: }~\text{Show that the points }\widehat{i}-\widehat{j}+3\widehat{k}\text{ and }3\widehat{i}+3\widehat{j}+3\widehat{k} \\ \text{are equidistant from the plane }\overrightarrow{r}\cdot(5\widehat{i}+2\widehat{j}-7\widehat{k})+9=0.
\displaystyle \text{Answer:}
\displaystyle \text{The given plane is}
\displaystyle \vec r\cdot(5\hat i+2\hat j-7\hat k)+9=0.
\displaystyle \Rightarrow \vec r\cdot(-5\hat i-2\hat j+7\hat k)=9.
\displaystyle \text{We know that the perpendicular distance }p\text{ of a point }P\text{ of position vector }\vec a\text{ from the plane }\vec r\cdot\vec n=d\text{ is given by}
\displaystyle p=\frac{\left|\vec a\cdot\vec n-d\right|}{\left|\vec n\right|}.
\displaystyle \text{Finding the distance from }\hat i-\hat j+3\hat k\text{ to the given plane}
\displaystyle \text{Here, }\vec a=\hat i-\hat j+3\hat k,\ \vec n=-5\hat i-2\hat j+7\hat k,\ d=9.
\displaystyle \text{So, the required distance }p\text{ is given by}
\displaystyle p=\frac{\left|(\hat i-\hat j+3\hat k)\cdot(-5\hat i-2\hat j+7\hat k)-9\right|}{\sqrt{25+4+49}}.
\displaystyle =\frac{\left|-5+2+21-9\right|}{\sqrt{78}}.
\displaystyle =\frac{9}{\sqrt{78}}\ \text{units.}\qquad(1)
\displaystyle \text{Finding the distance from }3\hat i+3\hat j+3\hat k\text{ to the given plane}
\displaystyle \text{Here, }\vec a=3\hat i+3\hat j+3\hat k,\ \vec n=-5\hat i-2\hat j+7\hat k,\ d=9.
\displaystyle \text{So, the required distance }p\text{ is given by}
\displaystyle p=\frac{\left|(3\hat i+3\hat j+3\hat k)\cdot(-5\hat i-2\hat j+7\hat k)-9\right|}{\sqrt{25+4+49}}.
\displaystyle =\frac{\left|-15-6+21-9\right|}{\sqrt{78}}.
\displaystyle =\frac{9}{\sqrt{78}}\ \text{units.}\qquad(2)
\displaystyle \text{From (1) and (2), we can say that the given points are equidistant from the given plane.}

\displaystyle \textbf{Question 3: }~\text{Find the distance of the point }(2,3,-5)\text{ from the plane } \\ x+2y-2z-9=0.
\displaystyle \text{Answer:}
\displaystyle \text{We know that the distance of the point }(x_1,y_1,z_1)\text{ from the plane }ax+by+cz+d=0\text{ is given by}
\displaystyle \frac{\left|ax_1+by_1+cz_1+d\right|}{\sqrt{a^2+b^2+c^2}}.
\displaystyle \text{So, the required distance}
\displaystyle =\frac{\left|(2)+2(3)-2(-5)-9\right|}{\sqrt{1^2+2^2+(-2)^2}}.
\displaystyle =\frac{\left|2+6+10-9\right|}{\sqrt{1+4+4}}.
\displaystyle =\frac{9}{3}.
\displaystyle =3\ \text{units.}

\displaystyle \textbf{Question 4: }~\text{Find the equations of the planes parallel to the plane } \\ x+2y-2z+8=0\text{ which are at distance of }2\text{ units from the point }(2,1,1).
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane parallel to the given plane is}
\displaystyle x+2y-2z+k=0.\qquad(1)
\displaystyle \text{It is given that plane (1) is at a distance of }2\text{ units from }(2,1,1).
\displaystyle \Rightarrow \frac{\left|2+2-2+k\right|}{\sqrt{1^2+2^2+(-2)^2}}=2.
\displaystyle \Rightarrow \frac{\left|2+k\right|}{3}=2.
\displaystyle \Rightarrow |2+k|=6.
\displaystyle \Rightarrow 2+k=6\ \text{or}\ 2+k=-6.
\displaystyle \Rightarrow k=4\ \text{or}\ k=-8.
\displaystyle \text{Substituting these two values one by one in (1), we get}
\displaystyle x+2y-2z+4=0\ \text{and}
\displaystyle x+2y-2z-8=0,
\displaystyle \text{which are the equations of the required planes.}

\displaystyle \textbf{Question 5: }~\text{Show that the points }(1,1,1)\text{ and }(-3,0,1) \\ \text{ are equidistant from the plane} 3x+4y-12z+13=0.
\displaystyle \text{Answer:}
\displaystyle \text{We know that the distance of the point }(x_1,y_1,z_1)\text{ from the plane }ax+by+cz+d=0\text{ is given by}
\displaystyle \frac{\left|ax_1+by_1+cz_1+d\right|}{\sqrt{a^2+b^2+c^2}}.
\displaystyle \text{Distance of the point }(1,1,1)\text{ from the plane }3x+4y-12z+13=0
\displaystyle \text{The required distance}
\displaystyle =\frac{\left|3(1)+4(1)-12(1)+13\right|}{\sqrt{3^2+4^2+(-12)^2}}.
\displaystyle =\frac{\left|3+4-12+13\right|}{\sqrt{9+16+144}}.
\displaystyle =\frac{8}{13}\ \text{units.}\qquad(1)
\displaystyle \text{Distance of the point }(-3,0,1)\text{ from the plane }3x+4y-12z+13=0
\displaystyle \text{The required distance}
\displaystyle =\frac{\left|3(-3)+4(0)-12(1)+13\right|}{\sqrt{3^2+4^2+(-12)^2}}.
\displaystyle =\frac{\left|-9+0-12+13\right|}{\sqrt{9+16+144}}.
\displaystyle =\frac{8}{13}\ \text{units.}\qquad(2)
\displaystyle \text{From (1) and (2), we can say that the given points are equidistant from the given plane.}

\displaystyle \textbf{Question 6: }~\text{Find the equations of the planes parallel to the plane } x-2y+2z-3=0\text{ and which are at a unit distance from the point } (1,1,1).
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane parallel to the given plane is}
\displaystyle x-2y+2z+k=0.\qquad(1)
\displaystyle \text{It is given that plane (1) is at a distance of }1\text{ unit from }(1,1,1).
\displaystyle \Rightarrow \frac{\left|1-2+2+k\right|}{\sqrt{1^2+(-2)^2+2^2}}=1.
\displaystyle \Rightarrow \frac{\left|1+k\right|}{3}=1.
\displaystyle \Rightarrow |1+k|=3.
\displaystyle \Rightarrow 1+k=3\ \text{or}\ 1+k=-3.
\displaystyle \Rightarrow k=2\ \text{or}\ k=-4.
\displaystyle \text{Substituting these two values one by one in (1), we get}
\displaystyle x-2y+2z+2=0\ \text{and}
\displaystyle x-2y+2z-4=0,
\displaystyle \text{which are the equations of the required planes.}

\displaystyle \textbf{Question 7: }~\text{Find the distance of the point }(2,3,5)\text{ from the }xy\text{-plane.}
\displaystyle \text{Answer:}
\displaystyle \text{We know that the distance of the point }(x_1,y_1,z_1)\text{ from the plane }ax+by+cz+d=0\text{ is given by}
\displaystyle \frac{\left|ax_1+by_1+cz_1+d\right|}{\sqrt{a^2+b^2+c^2}}.
\displaystyle \text{Equation of the }xy\text{-plane is }z=0,\ \text{which means }0x+0y+z+0=0.
\displaystyle \text{So, the required distance}
\displaystyle =\frac{\left|0(2)+0(3)+1(5)+0\right|}{\sqrt{0^2+0^2+1^2}}.
\displaystyle =\frac{5}{1}.
\displaystyle =5\ \text{units.}

\displaystyle \textbf{Question 8: }~\text{Find the distance of the point }(3,3,3)\text{ from the plane }\overrightarrow{r}\cdot(5\widehat{i}+2\widehat{j}-7\widehat{k})+9=0.
\displaystyle \text{Answer:}
\displaystyle \text{The given plane is}
\displaystyle \vec r\cdot(5\hat i+2\hat j-7\hat k)+9=0.
\displaystyle \Rightarrow \vec r\cdot(-5\hat i-2\hat j+7\hat k)=9.
\displaystyle \text{We know that the perpendicular distance }p\text{ of a point }P\text{ of position vector }\vec a\text{ from the plane }\vec r\cdot\vec n=d\text{ is given by}
\displaystyle p=\frac{\left|\vec a\cdot\vec n-d\right|}{\left|\vec n\right|}.
\displaystyle \text{Finding the distance from }(3,3,3)\text{ which means }3\hat i+3\hat j+3\hat k\text{ to the given plane}
\displaystyle \text{Here, }\vec a=3\hat i+3\hat j+3\hat k,\ \vec n=-5\hat i-2\hat j+7\hat k,\ d=9.
\displaystyle \text{So, the required distance }p
\displaystyle =\frac{\left|(3\hat i+3\hat j+3\hat k)\cdot(-5\hat i-2\hat j+7\hat k)-9\right|}{\sqrt{25+4+49}}.
\displaystyle =\frac{\left|-15-6+21-9\right|}{\sqrt{78}}.
\displaystyle =\frac{9}{\sqrt{78}}\ \text{units.}

\displaystyle \textbf{Question 9: }~\text{If the product of distances of the point }(1,1,1)\text{ from the origin and the plane } \\ x-y+z+\lambda=0\text{ be }5,\text{ find the value of }\lambda.
\displaystyle \text{Answer:}

\displaystyle \text{Given point }(1,1,1)
\displaystyle \text{Distance from origin }=\sqrt{1^2+1^2+1^2}
\displaystyle =\sqrt{3}
\displaystyle \text{Given plane }x-y+z+\lambda=0
\displaystyle \text{Distance of point }(1,1,1)\text{ from plane }Ax+By+Cz+D=0
\displaystyle =\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}
\displaystyle A=1,\ B=-1,\ C=1,\ D=\lambda
\displaystyle x_1=1,\ y_1=1,\ z_1=1
\displaystyle \text{Distance from plane }=\frac{|1-1+1+\lambda|}{\sqrt{1^2+(-1)^2+1^2}}
\displaystyle =\frac{|1+\lambda|}{\sqrt{3}}
\displaystyle \text{Product of distances }=\sqrt{3}\cdot\frac{|1+\lambda|}{\sqrt{3}}
\displaystyle =|1+\lambda|
\displaystyle \text{Given }|1+\lambda|=5
\displaystyle 1+\lambda=\pm5
\displaystyle \lambda=4\ \text{or}\ \lambda=-6

\displaystyle \textbf{Question 10: }~\text{Find an equation for the set of all points that are equidistant from} \\ \text{the planes }3x-4y+12z=6\text{ and }4x+3z=7.
\displaystyle \text{Answer:}
\displaystyle \text{Let }(x,y,z)\text{ be a point which is equidistant from the given planes. Then,}
\displaystyle \frac{|3x-4y+12z-6|}{\sqrt{9+16+144}}=\frac{|4x+3z-7|}{\sqrt{16+9}}.
\displaystyle \Rightarrow \frac{3x-4y+12z-6}{13}=\pm\frac{4x+3z-7}{5}.
\displaystyle \Rightarrow 15x-20y+60z-30=52x+39z-91.
\displaystyle \text{or }15x-20y+60z-30=-52x-39z+91.
\displaystyle \Rightarrow 37x+20y-21z-61=0.
\displaystyle \text{or }67x-20y+99z=121.

\displaystyle \textbf{Question 11: }~\text{Find the distance between the point }(7,2,4) \text{ and the plane} \\ \text{determined by the points }A(2,5,-3),\ B(-2,-3,5)\text{ and }(5,3,-3).\ \text{[CBSE\ 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{The given points are }A(2,5,-3),B(-2,-3,5)\text{ and }C(5,3,-3).
\displaystyle \text{The equation of the plane }ABC\text{ is given by}
\displaystyle \begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix}=0
\displaystyle \Rightarrow \begin{vmatrix}x-2&y-5&z+3\\-2-2&-3-5&5+3\\5-2&3-5&-3+3\end{vmatrix}=0
\displaystyle \Rightarrow \begin{vmatrix}x-2&y-5&z+3\\-4&-8&8\\3&-2&0\end{vmatrix}=0
\displaystyle \Rightarrow \begin{vmatrix}x-2&y-5&z+3\\1&2&-2\\3&-2&0\end{vmatrix}=0
\displaystyle \Rightarrow -4(x-2)-6(y-5)-8(z+3)=0
\displaystyle \Rightarrow 2(x-2)+3(y-5)+4(z+3)=0
\displaystyle \Rightarrow 2x+3y+4z-7=0
\displaystyle \therefore \text{Distance between the point }(7,2,4)\text{ and the plane }2x+3y+4z-7=0
\displaystyle =\frac{|2\times7+3\times2+4\times4-7|}{\sqrt{2^2+3^2+4^2}}
\displaystyle =\frac{|14+6+16-7|}{\sqrt{4+9+16}}
\displaystyle =\frac{29}{\sqrt{29}}
\displaystyle =\sqrt{29}\text{ units}

\displaystyle \textbf{Question 12: }~\text{A plane makes intercepts }-6,\ 3,\ 4\text{ respectively on the coordinate axes.} \\ \text{Find the length of the perpendicular from the origin on it. \ [CBSE\ 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{We know that the equation of the plane which makes intercepts }a,b,c\text{ on the coordinate axes is}
\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
\displaystyle \text{So, the equation of the plane which makes intercepts }-6,3,4\text{ on the }x\text{-axis, }y\text{-axis and }z\text{-axis respectively is}
\displaystyle \frac{x}{-6}+\frac{y}{3}+\frac{z}{4}=1
\displaystyle \Rightarrow -2x+4y+3z=12
\displaystyle \Rightarrow 2x-4y-3z+12=0
\displaystyle \therefore \text{Length of the perpendicular from }(0,0,0)\text{ to the plane }2x-4y-3z+12=0
\displaystyle =\frac{|2\times0-4\times0-3\times0+12|}{\sqrt{2^2+(-4)^2+(-3)^2}}
\displaystyle =\frac{12}{\sqrt{4+16+9}}
\displaystyle =\frac{12}{\sqrt{29}}\text{ units}
\displaystyle \therefore \text{The length of the perpendicular from the origin to the plane is }\frac{12}{\sqrt{29}}\text{ units.}

\displaystyle \textbf{Question 13: }~\text{Find the distance of the point }(1,-2,4)\text{ from plane passing} \\ \text{through the point }(1,2,2)\text{ and perpendicular to the planes }x-y+2z=3\text{ and } \\ 2x-2y+z+12=0.\ \text{[CBSE\ 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of plane passing through the point }(1,2,2)\text{ be}
\displaystyle a(x-1)+b(y-2)+c(z-2)=0\qquad ...(1)
\displaystyle \text{Here, }a,b,c\text{ are the direction ratios of the normal to the plane.}
\displaystyle \text{The equations of the given planes are }x-y+2z=3\text{ and }2x-2y+z+12=0.
\displaystyle \text{Plane (1) is perpendicular to the given planes.}
\displaystyle a-b+2c=0\qquad ...(2)
\displaystyle 2a-2b+c=0\qquad ...(3)
\displaystyle \text{Eliminating }a,b,c\text{ from (1),(2) and (3), we get}
\displaystyle \begin{vmatrix}x-1&y-2&z-2\\1&-1&2\\2&-2&1\end{vmatrix}=0
\displaystyle \Rightarrow (x-1)\begin{vmatrix}-1&2\\-2&1\end{vmatrix}-(y-2)\begin{vmatrix}1&2\\2&1\end{vmatrix}+(z-2)\begin{vmatrix}1&-1\\2&-2\end{vmatrix}=0
\displaystyle \Rightarrow (x-1)(-1-4)-(y-2)(1-4)+(z-2)(-2+2)=0
\displaystyle \Rightarrow -5(x-1)+3(y-2)+0=0
\displaystyle \Rightarrow 5x-3y-5-6=0
\displaystyle \Rightarrow 5x-3y-11=0
\displaystyle \Rightarrow x+y-3=0
\displaystyle \therefore \text{Distance of the point }(1,-2,4)\text{ from the plane }x+y-3=0
\displaystyle =\frac{|1-2-3|}{\sqrt{1^2+1^2+0^2}}
\displaystyle =\frac{|-4|}{\sqrt{2}}
\displaystyle =2\sqrt{2}\text{ units}

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