Question

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

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

Answer 1

The correct option is D

$2\sqrt{2}$

Let the equation of plane be
$a(x-1)+b(y+2)+c(z-1)=0$
Which is perpendicular to $2x-2y+z=0andx-y+2z=4\Rightarrow 2a-2b+c=0anda-b+2c=0\Rightarrow \frac{a}{-3}=\frac{b}{-3}=\frac{c}{0}\Rightarrow \frac{a}{1}=\frac{b}{1}=\frac{c}{0}$
So, the equation of plane is $x-1+y+2=0=>x+y+1=0$
Its distance from the point (1, 2, 2) is
$\frac{|1+2+1|}{\sqrt{2}}=2\sqrt{2}$