11. A line $l$ with a slope of $\frac{1}{3}$ intersects the ellipse $C: \frac{x^{2}}{36}+\frac{y^{2}}{4}=1$ at points $A$ and $B$ (as shown in the figure), and $P(3 \sqrt{2}, \sqrt{2})$ is above and to the left of line $l$.
(1) Prove that the incenter of $\triangle P A B$ lies on a fixed line;
(2) If $\angle A P B=60^{\circ}$, find the area of $\triangle P A B$.
Answer

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$$
\left\{\begin{array}{l}
x^{2}+9 y^{2}=36 \\
3 y=x+3 m
\end{array}\right.
$$
(1) Let the line $l: y=\frac{1}{3} x+m \Rightarrow 3 y=x+3 m$, and combine the equations
$$
\Rightarrow 2 x^{2}+6 m x+9 m^{2}-36=0 \text {. }
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then $x_{1}+x_{2}=-3 m, x_{1} x_{2}=\frac{9 m^{2}-36}{2}$.
Thus, $k_{P A}+k_{P B}=\frac{y_{1}-\sqrt{2}}{x_{1}-3 \sqrt{2}}+\frac{y_{2}-\sqrt{2}}{x_{2}-3 \sqrt{2}}$
$$
\begin{array}{l}
=\frac{\left(3 y_{1}-3 \sqrt{2}\right)\left(x_{2}-3 \sqrt{2}\right)+\left(3 y_{2}-3 \sqrt{2}\right)\left(x_{1}-3 \sqrt{2}\right)}{3\left(x_{1}-3 \sqrt{2}\right)\left(x_{2}-3 \sqrt{2}\right)} . \\
\text { Since }\left(3 y_{1}-3 \sqrt{2}\right)\left(x_{2}-3 \sqrt{2}\right)+\left(3 y_{2}-3 \sqrt{2}\right)\left(x_{1}-3 \sqrt{2}\right) \\
=\left(x_{1}+3 m-3 \sqrt{2}\right)\left(x_{2}-3 \sqrt{2}\right)+\left(x_{2}+3 m-3 \sqrt{2}\right)\left(x_{1}-3 \sqrt{2}\right) \\
=2 x_{1} x_{2}+(3 m-6 \sqrt{2})\left(x_{1}+x_{2}\right)-6 \sqrt{2}(3 m-3 \sqrt{2}) \\
=9 m^{2}-36-3 m(3 m-6 \sqrt{2})-6 \sqrt{2}(3 m-3 \sqrt{2})=0 \Rightarrow k_{P A}+k_{P B}=0,
\end{array}
$$
$=9 m^{2}-36-3 m(3 m-6 \sqrt{2})-6 \sqrt{2}(3 m-3 \sqrt{2})=0 \Rightarrow k_{P A}+k_{P B}=0$, hence $P A, P B$ are symmetric about the line $x=3 \sqrt{2}$. Since the line $l$ is below point $P$, the line $x=3 \sqrt{2}$ is the angle bisector of $\angle A P B$. Therefore, the incenter of $\triangle P A B$ lies on the fixed line $x=3 \sqrt{2}$.
$$
\begin{array}{l}
\text { (2) } \angle A P B=60^{\circ} \Rightarrow k_{P A} k_{P B}=\frac{\left(y_{1}-\sqrt{2}\right)\left(y_{2}-\sqrt{2}\right)}{\left(x_{1}-3 \sqrt{2}\right)\left(x_{2}-3 \sqrt{2}\right)}=-3 \\
\Rightarrow \frac{\left(3 y_{1}-3 \sqrt{2}\right)\left(3 y_{2}-3 \sqrt{2}\right)}{\left(x_{1}-3 \sqrt{2}\right)\left(x_{2}-3 \sqrt{2}\right)}=\frac{\left(x_{1}+3 m-3 \sqrt{2}\right)\left(x_{2}+3 m-3 \sqrt{2}\right)}{\left(x_{1}-3 \sqrt{2}\right)\left(x_{2}-3 \sqrt{2}\right)} \\
=\frac{x_{1} x_{2}+(3 m-3 \sqrt{2})\left(x_{1}+x_{2}\right)+(3 m-3 \sqrt{2})^{2}}{x_{1} x_{2}-3 \sqrt{2}\left(x_{1}+x_{2}\right)+18}=\frac{\frac{9 m^{2}}{2}-9 m \sqrt{2}}{\frac{9 m^{2}}{2}+9 m \sqrt{2}} \\
=\frac{m-2 \sqrt{2}}{m+2 \sqrt{2}}=-27 \Rightarrow m=-\frac{13 \sqrt{2}}{7} \Rightarrow l: y=\frac{1}{3} x-\frac{13 \sqrt{2}}{7} .
\end{array}
$$

On the line $l$, let $y=\sqrt{2} \Rightarrow M\left(\frac{60 \sqrt{2}}{7}, \sqrt{2}\right) \Rightarrow|P M|=\frac{39 \sqrt{2}}{7}$.
Thus, $S_{\triangle P A B}=S_{\triangle P M B}-S_{\triangle P M A}=\frac{1}{2}|P M| \cdot\left|y_{1}-y_{2}\right|=\frac{1}{6}|P M| \cdot\left|x_{1}-x_{2}\right|$ $=\frac{1}{12}|P M| \sqrt{36 m^{2}-8\left(9 m^{2}-36\right)}=\frac{1}{2}|P M| \sqrt{8-m^{2}}$
$=\frac{1}{2} \cdot \frac{39 \sqrt{2}}{7} \sqrt{8-\left(-\frac{13 \sqrt{2}}{7}\right)^{2}}=\frac{1}{2} \cdot \frac{39 \sqrt{2}}{7} \cdot \frac{3 \sqrt{6}}{7}=\frac{117 \sqrt{3}}{49}$.