$The dis\tan ce of the z-axis from the image of the point M (2,-3,3) in the plane x-2y-z+1=0, is$

$\begin{array}{l}If the image of point \left(2,-3,3\right) in the plane x-2y-z+1=0 is \left(a,b,c\right) then \text{}we have\text{}\\ \frac{a-2}{1}=\frac{b+3}{-2}=\frac{c-3}{-1}=\frac{-2\left(2-2\left(-3\right)-3+1\right)}{1^2+{\left(-2\right)}^2+{\left(-1\right)}^2}=-2\text{}\\ Hence the image is \left(0,1,5\right)\text{}\\ Dis\tan ce of image of the point from z-axis is \sqrt{x^2+y^2}=1\end{array}$