Let P be a point in the first octant whose image Q in the plane x + y = 3 (that is, the line segment PQ perpendicular to the plane x + y = 3 and the mid-point of PQ lies in the plane x + y = 3) lies on the z-axis. Let the distance of P from the x-axis be 5. If R is the image of P in the x y - plane, then the length of PR is

Let P($\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$) be a point in the first octant. The image of P in the xy plane is the point
R($\left(\mathrm{\alpha },\mathrm{\beta },-\mathrm{\gamma }\right)$) .
The coordinates of the image of P($\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$) in the plane x + y + 0z - 3 = 0 are given by
$\begin{array}{l}\frac{\mathrm{x}-\mathrm{\alpha }}{1}=\frac{\mathrm{y}-\mathrm{\beta }}{1}=\frac{\mathrm{z}-\mathrm{\gamma }}{0}=-2\frac{(\mathrm{\alpha }+\mathrm{\beta }+0\mathrm{\gamma }-3)}{\left(1^2+1^2+0^2\right)}\\ \Rightarrow \mathrm{x}=3-\mathrm{\beta },\mathrm{y}=3-\mathrm{\alpha },\mathrm{z}=\mathrm{\gamma }\end{array}$
So, The coordinates of Q are (3 $-\mathrm{\beta }$,3 -$\mathrm{\alpha },\mathrm{\gamma }$ ). It lies on the z-axis.
Therefore, $\mathrm{\beta }=3,\mathrm{\alpha }=3$.Thus, the coordinates of P are (3,3,$\mathrm{\gamma }$). Its distance form x axis
is 5.
$\therefore \sqrt{9+{\mathrm{\gamma }}^2}=5\Rightarrow \mathrm{\gamma }=4$
Thus, the coordinates of P and R are (3, 3, 4) and (3, 3, -4).
Hence, PR = 8