11. In $\triangle A B C$, the sides opposite to the internal angles $A, B, C$ are $a, b, c$ respectively. It is known that the area of $\triangle A B C$ is $\frac{a^{2}}{3 \sin A}$.
(1) Find $\sin B \sin C$;
(2) If $6 \cos B \cos C=1, a=3$, find the perimeter of $\triangle A B C$.
(1) $S_{\triangle A B C}=2 R^{2} \sin B \sin C \sin A=\frac{a^{2}}{3 \sin A} \Rightarrow \sin B \sin C=\frac{a^{2}}{2 R^{2} \cdot 3 \sin ^{2} A}=\frac{4 R^{2} \sin ^{2} A}{6 R^{2} \sin ^{2} A}=\frac{2}{3}$.
(2) $\cos B \cos C=\frac{1}{6}, \sin B \sin C=\frac{2}{3} \Rightarrow \cos (B+C)=-\frac{1}{2} \Rightarrow A=60^{\circ}$,

Thus $2 R=\frac{a}{\sin A}=2 \sqrt{3} \Rightarrow b c=4 R^{2} \sin B \sin C=8$,
Also $a^{2}=b^{2}+c^{2}-b c=(b+c)^{2}-3 b c \Rightarrow(b+c)^{2}=24+9=33 \Rightarrow a+b+c=3+\sqrt{33}$.
Therefore, the perimeter of $\triangle A B C$ is $3+\sqrt{33}$.