11、(20 points) Given the hyperbola
$$
C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$
$F_{1}, F_{2}$ are the left and right foci of the hyperbola $C$, respectively, and $P$ is a point on the right branch of the hyperbola $C$ such that $\angle F_{1} P F_{2}=\frac{\pi}{3}$. The area of $\triangle F_{1} P F_{2}$ is $3 \sqrt{3} a^{2}$.
(1) Find the eccentricity $e$ of the hyperbola $C$.
(2) Let $A$ be the left vertex of the hyperbola $C$, and $Q$ be any point on the hyperbola $C$ in the first quadrant. Does there exist a constant $\lambda(\lambda>0)$ such that $\angle Q F_{2} A=\lambda \angle Q A F_{2}$ always holds? If it exists, find the value of $\lambda$; if not, explain the reason.
(1) From $S_{\triangle F_{1} P F_{2}}=\frac{1}{2}\left|P F_{1}\right|\left|P F_{2}\right| \sin \frac{\pi}{3}=3 \sqrt{3} a^{2}$, we get $\left|P F_{1}\right|\left|P F_{2}\right|=12 a^{2}$.
By the cosine rule,
$$
\begin{array}{c}
4 c^{2}=\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}-2\left|P F_{1}\right|\left|P F_{2}\right| \cos \frac{\pi}{3}=\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}-\left|P F_{1}\right|\left|P F_{2}\right| \\
=\left(\left|P F_{1}\right|-\left|P F_{2}\right|\right)^{2}+\left|P F_{1}\right|\left|P F_{2}\right|=16 a^{2}
\end{array}
$$

Thus, $e^{2}=4, e=2$.
(5 points)
(2) $c=2 a$. The hyperbola $C$ is $3 x^{2}-y^{2}=3 a^{2}, A F_{2}=3 a$.
Let $\angle Q F_{2} A=\alpha, \angle Q A F_{2}=\beta, Q(x, y)$
When $\alpha=\frac{\pi}{2}$, from $\left|Q F_{2}\right|=\left|A F_{2}\right|=3 a$, we have $\beta=\frac{\pi}{4}, \alpha=2 \beta$.
(10 points)
When $\alpha \neq \frac{\pi}{2}$, $\tan \alpha=\frac{y}{2 a-x}, \tan \beta=\frac{y}{a+x}$
$$
\begin{array}{c}
\tan 2 \beta=\frac{\frac{2 y}{a+x}}{1-\left(\frac{y}{a+x}\right)^{2}}=\frac{2 y(a+x)}{(a+x)^{2}-y^{2}}=\frac{2 y(a+x)}{(a+x)^{2}+3 a^{2}-3 x^{2}} \\
=\frac{2 y}{(a+x)+3(a-x)}=\frac{2 y}{2 a-x}=\tan \alpha
\end{array}
$$

Therefore, $\alpha=2 \beta$.
(20 points)