11. The number of tetrahedra that can be formed with the vertices of a regular hexagonal prism is $\qquad$.

 untranslated part: $\qquad$
11.426.
Among 12 vertices, choosing any 4, there are $\mathrm{C}_{12}^{4}$ different ways, but when the 4 chosen points are coplanar, they cannot form a tetrahedron and should be excluded.
The 4 chosen points being coplanar can be divided into 4 cases:
(1) 4 points are on the same side, with 6 sides, and the side is a rectangle, so there are 6 ways;
(2) 4 points are on the upper base or the lower base, with a total of $2 C_{6}^{4}$ ways;
(3) 4 points are on the diagonal plane perpendicular to the upper and lower bases, with a total of $\mathrm{C}_{6}^{2}-6=9$ ways;
(4) 4 points are coplanar and on the diagonal plane not perpendicular to the upper and lower bases, with a total of $2 \times 6+2 \times 3 \times 2=24$ ways.
Therefore, the number of tetrahedra is
$$
\begin{array}{l}
\mathrm{C}_{12}^{4}-6-2 \mathrm{C}_{6}^{4}-9-24 \\
=495-6-2 \times 15-9-24=426 .
\end{array}
$$