In $\triangle A B C$ , $\sin ^ { 2 } A - \sin ^ { 2 } B - \sin ^ { 2 } C = \sin B \sin C$ . (1) Find $A$ ; (2) If $B C = 3$ , find the maximum value of the perimeter of $\triangle A B C$ .
(1) By the law of sines and the given condition, we have $BC^2 - AC^2 - AB^2 = AC \cdot AB$. By the law of cosines, $BC^2 = AC^2 + AB^2 - 2AC \cdot AB \cos A$. From these, we get $\cos A = -\frac{1}{2}$. Since $0 < A < \pi$, we have $A = \frac{2\pi}{3}$. (2) By the law of sines and (1), $\frac{AC}{\sin B} = \frac{AB}{\sin C} = \frac{BC}{\sin A} = 2\sqrt{3}$, thus $AC = 2\sqrt{3}\sin B$, $AB = 2\sqrt{3}\sin(\pi - A - B) = 3\cos B - \sqrt{3}\sin B$. Therefore, $BC + AC + AB = 3 + \sqrt{3}\sin B + 3\cos B = 3 + 2\sqrt{3}\sin\left(B + \frac{\pi}{3}\right)$. Since $0 < B < \frac{\pi}{3}$, the perimeter reaches its maximum value $3 + 2\sqrt{3}$ when $B = \frac{\pi}{6}$.
In $\triangle A B C$ , $\sin ^ { 2 } A - \sin ^ { 2 } B - \sin ^ { 2 } C = \sin B \sin C$ .
(1) Find $A$ ;
(2) If $B C = 3$ , find the maximum value of the perimeter of $\triangle A B C$ .