3. The three sides of $\triangle A B C$, $a$, $b$, and $c$, satisfy $b+c=$ $8, bc=a^{2}-12a+52$. Then the perimeter of $\triangle A B C$ is ( ).
(A) 10
(B) 14
(C) 16
(D) cannot be determined
3. (B).

Since $(b-c)^{2}=64-4\left(a^{2}-12 a+52\right)$
$$
=-4(a-6)^{2} \geqslant 0 \text {, }
$$

Therefore, $a=6$, and $b=c$.
Also, $b+c=8$, then $b=c=4$. Thus, $a+b+c=14$.