11 (20 points) 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 $\triangle P A B$.
(111) (1) Let the line $l: y=\frac{1}{3} x+m, A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Substitute $y=\frac{1}{3} x+m$ into $\frac{x^{2}}{36}+\frac{y^{2}}{4}=1$, and simplify to get
$$
2 x^{2}+6 m x+9 m^{2}-36=0.
$$

Thus, we have $x_{1}+x_{2}=-3 m, x_{1} x_{2}=\frac{9 m^{2}-36}{2}, k_{P A}=\frac{y_{1}-\sqrt{2}}{x_{1}-3 \sqrt{2}}$,
$k_{P B}=\frac{y_{2}-\sqrt{2}}{x_{2}-3 \sqrt{2}}$. Then
$$
\begin{aligned}
k_{P A}+k_{P B} & =\frac{y_{1}-\sqrt{2}}{x_{1}-3 \sqrt{2}}+\frac{y_{2}-\sqrt{2}}{x_{2}-3 \sqrt{2}} \\
& =\frac{\left(y_{1}-\sqrt{2}\right)\left(x_{2}-3 \sqrt{2}\right)+\left(y_{2}-\sqrt{2}\right)\left(x_{1}-3 \sqrt{2}\right)}{\left(x_{1}-3 \sqrt{2}\right)\left(x_{2}-3 \sqrt{2}\right)},
\end{aligned}
$$

In the above expression,
$$
\begin{aligned}
\text { Numerator } & =\left(\frac{1}{3} x_{1}+m-\sqrt{2}\right)\left(x_{2}-3 \sqrt{2}\right)+\left(\frac{1}{3} x_{2}+m-\sqrt{2}\right)\left(x_{1}-3 \sqrt{2}\right) \\
& =\frac{2}{3} x_{1} x_{2}+(m-2 \sqrt{2})\left(x_{1}+x_{2}\right)-6 \sqrt{2}(m-\sqrt{2}) \\
& =\frac{2}{3} \cdot \frac{9 m^{2}-36}{2}+(m-2 \sqrt{2})(-3 m)-6 \sqrt{2}(m-\sqrt{2}) \\
& =3 m^{2}-12-3 m^{2}+6 \sqrt{2} m-6 \sqrt{2} m+12=0,
\end{aligned}
$$

Thus, $k_{P A}+k_{P B}=0$.
Since $P$ is in the upper left of the line $l$, the angle bisector of $\angle A P B$ is a vertical line, so the center of the incircle of $\triangle P A B$ lies on the line $x=3 \sqrt{2}$.
(2) If $\angle A P B=60^{\circ}$, then from (1) we know $k_{P A}=\sqrt{3}, k_{P B}=-\sqrt{3}$.
The equation of the line $P A$ is:
$$
y-\sqrt{2}=\sqrt{3}(x-3 \sqrt{2}),
$$

Substitute into $\frac{x^{2}}{36}+\frac{y^{2}}{4}=1$, and eliminate $y$ to get
$$
14 x^{2}+9 \sqrt{6}(1-3 \sqrt{3}) x+18(13-3 \sqrt{3})=0.
$$

Its roots are $x_{1}$ and $3 \sqrt{2}$, so
$$
x_{1} \cdot 3 \sqrt{2}=\frac{18(13-3 \sqrt{3})}{14},
$$

which gives
$$
x_{1}=\frac{3 \sqrt{2}(13-3 \sqrt{3})}{14}.
$$

Therefore,
$$
\begin{aligned}
|P A| & =\sqrt{1+(\sqrt{3})^{2}} \cdot\left|x_{1}-3 \sqrt{2}\right| \\
& =\frac{3 \sqrt{2}(3 \sqrt{3}+1)}{7}.
\end{aligned}
$$

Similarly, we can find
$$
|P B|=\frac{3 \sqrt{2}(3 \sqrt{3}-1)}{7}.
$$

Thus,
$$
\begin{aligned}
S_{\triangle P A B} & =\frac{1}{2} \cdot|P A| \cdot|P B| \cdot \sin 60^{\circ} \\
& =\frac{1}{2} \cdot \frac{3 \sqrt{2}(3 \sqrt{3}+1)}{7} \cdot \frac{3 \sqrt{2}(3 \sqrt{3}-1)}{7} \cdot \frac{\sqrt{3}}{2} \\
& =\frac{117 \sqrt{3}}{49}.
\end{aligned}
$$