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 Figure 1), and $P(3 \sqrt{2}, \sqrt{2})$ is located 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 $\triangle P A B$.
11. (1) Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,

line $l: y=\frac{1}{3} x+m$.
Substituting equation (1) into the equation of ellipse $C$ and simplifying, we get
$$
2 x^{2}+6 m x+9 m^{2}-36=0.
$$

Then $x_{1}+x_{2}=-3 m, x_{1} x_{2}=\frac{9 m^{2}-36}{2}$,
$$
k_{P A}=\frac{y_{1}-\sqrt{2}}{x_{1}-3 \sqrt{2}}, k_{P B}=\frac{y_{2}-\sqrt{2}}{x_{2}-3 \sqrt{2}}.
$$

Thus, $k_{P A}+k_{P B}$
$$
=\frac{\left(y_{1}-\sqrt{2}\right)\left(x_{2}-3 \sqrt{2}\right)+\left(y_{2}-\sqrt{2}\right)\left(x_{1}-3 \sqrt{2}\right)}{\left(x_{1}-3 \sqrt{2}\right)\left(x_{2}-3 \sqrt{2}\right)}.
$$

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

Therefore, $k_{P A}+k_{P B}=0$.
Since point $P$ is above and to the left of line $l$, the angle bisector of $\angle A P B$ is a vertical line.

Thus, the center of the incircle of $\triangle P A B$ lies on the line $x=3 \sqrt{2}$.
(2) If $\angle A P B=60^{\circ}$, from the conclusion in (1) we know
$$
k_{P A}=\sqrt{3}, k_{P B}=-\sqrt{3}.
$$

Substituting the line $l_{P A}: y-\sqrt{2}=\sqrt{3}(x-3 \sqrt{2})$ into the equation of ellipse $C$ and eliminating $y$, we get
$$
14 x^{2}+9 \sqrt{6}(1-3 \sqrt{3}) x+18(13-3 \sqrt{3})=0.
$$

Since the roots of the above equation are $x_{1}$ and $3 \sqrt{2}$, we have
$$
\begin{array}{l}
x_{1}=\frac{3 \sqrt{2}(13-3 \sqrt{3})}{14}. \\
\text { Then }|P A|=\sqrt{1+(\sqrt{3})^{2}}\left|x_{1}-3 \sqrt{2}\right| \\
=\frac{3 \sqrt{2}(3 \sqrt{3}+1)}{7}.
\end{array}
$$

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