$\text{ The distance of the }z\text{-axis from the image of the point }M(2,-3,3)\text{ in the plane }x-2y-z+1=0\text{, is }$

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

we have,

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

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