A mirror lies in the plane 2x - 2y + z = 3, with lengths in metres…

The image of $P$ in the plane $\vec{r} \cdot \vec{n} = k$ is $P' = \vec{p} - 2\,\dfrac{\vec{p} \cdot \vec{n} - k}{\lvert \vec{n} \rvert^2}\,\vec{n}$. The image in a line is $2F - P$. The shortest path $A \to M \to B$ off a mirror, with $A$ and $B$ on one side, is $A'B$.