11. Given
$$
\left(x+\sqrt{x^{2}+2002}\right)\left(y+\sqrt{y^{2}+2002}\right)=2002 .
$$

Then $x^{2}-3 x y-4 y^{2}-6 x-6 y+58=$ $\qquad$ .
11.58
$$
\begin{array}{l}
\text { Given }\left(x+\sqrt{x^{2}+2002}\right)\left(\sqrt{x^{2}+2002}-x\right) \\
=2002,
\end{array}
$$

we get $\sqrt{x^{2}+2002}-x=\sqrt{y^{2}+2002}+y$.
Similarly, $\sqrt{y^{2}+2002}-y=\sqrt{x^{2}+2002}+x$.
Adding these two equations gives $x+y=0$.
Therefore, $x^{2}-3 x y-4 y^{2}-6 x-6 y+58$
$$
=(x+y)(x-4 y)-6(x+y)+58=58 \text {. }
$$