3. Given the sides of $\triangle A B C$ are $a, b, c$ and satisfy the equations $a^{2}-a-2 b-2 c=0, a+2 b-2 c+3=0$. Then the largest interior angle of $\triangle A B C$ is ( ).
(A) $150^{\circ}$
(B) $120^{\circ}$
(C) $90^{\circ}$
(D) $60^{\circ}$
3. (B).

Given $\left\{\begin{array}{l}(a+2 b)-2 c=a^{2}, \\ (a+2 b)+2 c=-3 .\end{array}\right.$
Multiplying the two equations and simplifying, we get
$$
a^{2}+b^{2}-c^{2}=-a b<0 .
$$

Therefore, $\angle C$ is an obtuse angle, and $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=-\frac{1}{2}$. Hence, the largest interior angle is $\angle C=120^{\circ}$.