Question

A plane $E$ is perpendicular to the two planes $2x-2y+z=0$ and $x-y+2z=4$, and passes through the point $P(1,-1,1)$. If the distance of the plane $E$ from the point $Q(a,a,2)$ is $3\sqrt{2}$, then $(PQ{)}^2$ is equal to

• A. 9
• B. 12
• C. 21
• D. 33

Answer 1

Answer: (C)

Let equation of plane be $a(x-1)+b(y+1)+c(z-1)=0$.....(1)
It is perpendicular to the given two planes
$2a-2b+c=0$
$a-b+2c=0$
$\Rightarrow \frac{a}{3}=\frac{b}{3}=\frac{c}{0}$
Equation of plane be $x+y=0$
Now $\frac{|a+a|}{\sqrt{2}}=3\sqrt{2}\Rightarrow |2a|=6\Rightarrow a=\pm 3$
$P(3,3,2)$ or $P(-3,-3,2),Q(1,-1,1)$
$P{Q}^2=(3-1{)}^2+(3+1{)}^2+(2-1{)}^2=21$