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

Solution:   The equation of  the plane through the point $(1,-2,1)$

and  perpendicular to planes

$2x-2y+z=0 \hspace{0.5cm}and \hspace{0.5cm}x-y+2z=4$  is given by

$\begin{vmatrix} x-1 & y+2 & z-1\\ 2& -2 & 1\\ 1 &-1 &2 \end{vmatrix}=0\Rightarrow \hspace{0.5cm}x+y+1=0$

Its distance from the point    $(1,2,2)$   is    $\left | \frac{1+2+1}{\sqrt{2}} \right |=2\sqrt{2}.$