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 $(1,2,2)$ is

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

Answer 1

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

The equation of a plane passing through $(1,-2,1)$ is
$a(x-1)+b(y+2)+c(z-1)=0$ .........(i)
It is perpendicular to $2x-2y+z=0$ and $x-y+2z=4$.
Therefore, $2a-2b+c=0$ and $a-b+2c=0$
$\Rightarrow \frac{a}{-4+1}=\frac{-b}{4-1}=\frac{c}{-2+2}$
$\Rightarrow \frac{a}{-3}=\frac{b}{-3}=\frac{c}{0}$
On substituting the values of $a,b$ and $c$ in equation (i), we get
$-3(x-1)-3(y+2)+0(z-1)=0$
$\Rightarrow x+y+1=0$
Distance of this plane from $(1,2,2)$ is given by,
$d=\frac{1+2+1}{\sqrt{1+1}}=2\sqrt{2}$