jmc

All points on the line $x=-3$ are of the form $(-3,y)$, where $y$ is a real number. The distance from $(5,2)$ to $(-3,y)$ is $$\sqrt{(5-(-3){)}^2+(2-y{)}^2}$$ units. Setting this expression equal to 10, we find $$\begin{array}{rl}\sqrt{(5-(-3){)}^2+(2-y{)}^2}& =10\\ 64+(2-y{)}^2& =100\\ (2-y{)}^2& =36\\ 2-y& =\pm 6\\ y=2\pm 6.& \end{array}$$ The product of $2+6=8$ and $2-6=-4$ is $\boxed{-32}$.