Corners are sliced off a regular tetrahedron of edge length $3$ by planes that cut each edge at distances $1$ and $2$ from its ends, so that each of the four faces becomes a regular hexagon. What is the volume of the remaining solid?

1. A regular tetrahedron of edge $a$ has volume $\frac{a^3\sqrt2}{12}$. For $a=3$ this is $\frac{27\sqrt2}{12}=\frac{9\sqrt2}4$.
2. Each cut removes a small regular tetrahedron of edge $1$, of volume $\frac{\sqrt2}{12}$; there are $4$ of them, total $\frac{\sqrt2}3$.
3. The remaining volume is $\frac{9\sqrt2}4-\frac{\sqrt2}3=\boxed{\dfrac{23\sqrt2}{12}}$.