[Geometry]
Problem: The points $(1,2,3)$ and $(3,3,2)$ are vertices of a cube. Compute the product of all possible distinct volumes of the cube.

Solution: The distance between points $A(1,2,3)$ and $B(3,3,2)$ is $A B=\sqrt{(3-1)^{2}+(3-2)^{2}+(2-3)^{2}}=\sqrt{6}$. Denote by $s$ the side length of the cube. Consider three possibilities.

- If $\overline{A B}$ is an edge of the cube, then $A B=s$, so one possibility is $s_{1}=\sqrt{6}$.
- If $\overline{A B}$ is a face diagonal of the cube, then $A B=s \sqrt{2}$, so another possibility is $s_{2}=\sqrt{3}$.
- If $\overline{A B}$ is a space diagonal of the cube, then $A B=s \sqrt{3}$, so the last possibility is $s_{3}=\sqrt{2}$.

The answer is then $s_{1}^{3} s_{2}^{3} s_{3}^{3}=\left(s_{1} s_{2} s_{3}\right)^{3}=6^{3}=\mathbf{2 1 6}$.

Final answer: 216