1. $a, b, c$ are rational numbers, and the equation $a+b \sqrt{2}+c \sqrt{3}=\sqrt{5+2 \sqrt{6}}$ holds. Then the value of $2 a+995 b+1001 c$ is ( ).
(A) 1999
(B) 2000
(C) 2001
(D) cannot be determined
$-1 .(\mathrm{B})$
$5+2 \sqrt{6}=(\sqrt{3}+\sqrt{2})^{2}$. Therefore, $a=0, b=1, c=1$.