A plane passes through the point $(1, -2, 3)$ and is
parallel to the plane $2x - 2y + z = 0.$ The distance
of the point $(-1, 2, 0)$ from the plane is

The plane parallel to $2x-2y+z=0$ is

$2x-2y+z=k$

It passess through $(1,-2,3)\therefore k=2+4+3=9$

The plane is $2x-2y+z-9=0$

The distance of $(-1, 2, 0)$ from it is

$\frac{|-2-4-9|}{\sqrt{2^2+2^2+1}}=\frac{15}{3}=5$