11. Given the rectangular cuboid $A B C D-A_{1} B_{1} C_{1} D_{1}$ satisfies $A A_{1}=2, A D=3, A B=251$, the plane $A_{1} B D$ intersects $C C_{1} 、 C_{1} B_{1} 、 C_{1} D_{1}$ at points $L 、 M 、 N$. Then the volume of the tetrahedron $C_{1} L M N$ is $\qquad$ .
11.2008.

Since $B D / / M N, A_{1} D / / L M, A_{1} B / / L N$, therefore, $A_{1} 、 B 、 D$ are the midpoints of $M N 、 L M 、 L N$ respectively. Thus,
$$
\begin{array}{l}
C_{1} L=2 C C_{1}=4, C_{1} M=2 C_{1} B_{1}=6, \\
C_{1} N=2 C_{1} D_{1}=502 .
\end{array}
$$

Therefore, the volume of the tetrahedron $C_{1} L M N$ is
$$
\frac{1}{6} C_{1} L \cdot C_{1} M \cdot C_{1} N=2008 .
$$