There are two possible configurations. If $R L=12$, the side length of the square is $6 \sqrt{2}$. Now $$121=R K^{2}=R E^{2}+E K^{2}=(6 \sqrt{2})^{2}+E K^{2}$$ so $E K=7$. Then the possible values of $L K$ are $6 \sqrt{2} \pm 7$. Note that the area of $\triangle R L K$ is $$\frac{L K \cdot R E}{2}=L K \cdot 3 \sqrt{2}$$ and so the product of all possible areas are $$\begin{aligned} 3 \sqrt{2}(6 \sqrt{2}+7) \cdot 3 \sqrt{2}(6 \sqrt{2}-7) & =(6 \sqrt{2}+7)(6 \sqrt{2}-7) \cdot(3 \sqrt{2})^{2} \\ & =(72-49) \cdot 18=414 \end{aligned}$$