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

The equation of the plane E passing through the point P(1,-1,1) is 

a(x-1)+b(y+1)+c(z-1)=0. As E is perpendicular to the plane2x-2y+z=0

$\Rightarrow$2a-2b+c=0

Also E is perpendicular to the plane x-y+2z=4

$\Rightarrow$a-b+2c=0

By solving

$$\begin{gathered} \frac{a}{-4+1}=\frac{b}{1-4}=\frac{c}{0}=\lambda \\ \Rightarrow a=-3\lambda ,b=-3\lambda ,c=0 \\ Equation of the plane is \\ -3\lambda \left(x-1\right)-3\lambda \left(y+1\right)+0\left(z-1\right)=0 \\ \Rightarrow x+y=0 \\ The dis\tan ce of plane Ex+y=0 from pointQ\left(a,a,2\right) is 3\sqrt{2} \\ \Rightarrow \left|\frac{2a}{\sqrt{2}}\right|=3\sqrt{2}\Rightarrow a=3,-3 \\ \therefore Q\left(3,3,2\right) orQ\left(-3,-3,2\right) \\ \Rightarrow PQ=\sqrt{{\left(3-1\right)}^2+{\left(3+1\right)}^2+{\left(2-1\right)}^2}=\sqrt{4+16+1}=\sqrt{21} \\ \Rightarrow {\left(PQ\right)}^2=21 \end{gathered}$$