Find the distance of the z-axis from the image of the point M(2,-3,3) in the plane x-2y-z+1=0'

If image of point (2, - 3, 3) in the plane x - 2y - z + 1 = 0 is (a, b, c), then 

$\frac{\mathrm{a}-2}{1}=\frac{\mathrm{b}+3}{-2}=\frac{\mathrm{c}-3}{-1}=\frac{-2(2-2(-3)-3+1)}{(1{)}^2+(-2{)}^2+(-1{)}^2}=-2$

Hence, the image is (0, 1, 5) Obviously distance of image of the point from the z-axis is 1.