11 Because $B D // M N, A_{1} D // L M, A_{1} B // L N$, so $A_{1} 、 B 、 D$ are the midpoints of $M N 、 L M 、 L N$ respectively, thus we have $C_{1} L=2 C C_{1}=4, C_{1} M=2 C_{1} B_{1}=6$, $C_{1} N=2 C_{1} D_{1}=502$, from which we can obtain the volume of the tetrahedron $C_{1} L M N$ as
$$
\frac{1}{6} C_{1} L \cdot C_{1} M \cdot C_{1} N=2008
$$
(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 $C C_{1}$, $C_{1} B_{1}$, and $C_{1} D_{1}$ at points $L$, $M$, and $N$, respectively. Then the volume of the tetrahedron $C_{1} L M N$ is $\qquad$.