Question

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

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

Answer 1

Answer: (D)

$2\sqrt{2}$

General equation of a plane passes throgh $(1,-2,1)$ 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 (1) and (2), 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
$=d=\left|\frac{1+2+1}{\sqrt{1^2+1^2}}\right|=2\sqrt{2}$