Question

A plane passes through $(1,-2,1)$ and is perpendicular to the planes $2x-2y+z=0$ and $x-y+2z=4.$ Then the distance of the plane from the point $(1,2,2)$ is

• A. $0$
• B. $1$
• C. $\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

The correct option is D $2\sqrt{2}$

General equation of a plane passes through $(1,-2,1)$ is given by,

$a(x-1)+b(y+2)+c(z-1)=0$

Given this plane is perpendicular to given other two planes

$\Rightarrow 2a-2b+c=0...\left(1\right)$

and $a-b+2c=0....\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get $a=b,c=0$

Hence required plane is,

$x+y+1=0$

Therefore distance of point $(1,2,2)$ from this plane is

$=d=\left|\frac{1+2+1}{\sqrt{1^2+1^2}}\right|=2\sqrt{2}$