jmc

Extend $EA$ and $FB$ to meet at $G$, and $ED$ and $FC$ to meet at $H$. Now, we have a regular tetrahedron $EFGH$, which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s=12\sqrt{2}$. Using the formula for the volume of a regular tetrahedron, which is $V=\frac{\sqrt{2}{S}^3}{12}$, where S is the side length of the tetrahedron, the volume of our original solid is: $V=\frac{1}{2}\cdot \frac{\sqrt{2}\cdot (12\sqrt{2}{)}^3}{12}=\boxed{288}$.