Let $ABCD$ be a regular tetrahedron with edge length $6$, and let $E$ be any point inside it. What is the sum of the distances from $E$ to the four faces of the tetrahedron?

1. Join $E$ to the four vertices: this splits the tetrahedron into four pyramids with apex $E$, whose bases are the four faces. Let $\sigma$ be the area of a face and $d_1,\dots,d_4$ the distances from $E$ to the faces.
2. Then $V=\frac13\sigma(d_1+d_2+d_3+d_4)$. The same volume is $\frac13\sigma h$, where $h$ is the height of the tetrahedron, so $d_1+d_2+d_3+d_4=h$ for every $E$.
3. The height of a regular tetrahedron of edge $6$ is $\sqrt{6^2-(2\sqrt3)^2}=\sqrt{24}$, so the sum is $\boxed{2\sqrt6}$.