11. (20 points) Given the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{2}=1$, a moving line $l$ passing through the point $P\left(\frac{\sqrt{2}}{3},-\frac{1}{3}\right)$ but not through the point $Q(\sqrt{2}, 1)$ intersects the ellipse $C$ at points $A$ and $B$.
(1) Find $\angle A Q B$;
(2) Let the area of $\triangle Q A B$ be $S$, prove: $S<3$.
11. (1) If the slope of line $l$ exists, let its equation be $y=k x+b$.
Since point $P$ is on line $l$, we have,
$$
\begin{array}{l}
-\frac{1}{3}=\frac{\sqrt{2}}{3} k+b \\
\Rightarrow b=-\frac{1}{3}(\sqrt{2} k+1) .
\end{array}
$$

Combining the equation of line $l$ and the equation of ellipse $C$ and eliminating $y$ gives $\left(2 k^{2}+1\right) x^{2}+4 k b x+2 b^{2}-4=0$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l} 
x_{1}+x_{2}=-\frac{4 k b}{2 k^{2}+1} \\
x_{1} x_{2}=\frac{2 b^{2}-4}{2 k^{2}+1} \\
y_{1}+y_{2}=k\left(x_{1}+x_{2}\right)+2 b \\
=-\frac{4 k^{2} b}{2 k^{2}+1}+2 b=\frac{2 b}{2 k^{2}+1}, \\
y_{1} y_{2}=\left(k x_{1}+b\right)\left(k x_{2}+b\right) \\
=k^{2} x_{1} x_{2}+k b\left(x_{1}+x_{2}\right)+b^{2} \\
=k^{2} \cdot \frac{2 b^{2}-4}{2 k^{2}+1}+k b\left(-\frac{4 k b}{2 k^{2}+1}\right)+b^{2} \\
=\frac{b^{2}-4 k^{2}}{2 k^{2}+1} .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
\overrightarrow{Q A}=\left(x_{1}-\sqrt{2}, y_{1}-1\right), \\
\overrightarrow{Q B}=\left(x_{2}-\sqrt{2}, y_{2}-1\right) . \\
\text { Then } \overrightarrow{Q A} \cdot \overrightarrow{Q B} \\
=\left(x_{1}-\sqrt{2}\right)\left(x_{2}-\sqrt{2}\right)+\left(y_{1}-1\right)\left(y_{2}-1\right) \\
=x_{1} x_{2}-\sqrt{2}\left(x_{1}+x_{2}\right)+2+y_{1} y_{2}-\left(y_{1}+y_{2}\right)+1 \\
=\frac{2 b^{2}-4}{2 k^{2}+1}+\sqrt{2} \times \frac{4 k b}{2 k^{2}+1}+\frac{b^{2}-4 k^{2}}{2 k^{2}+1}- \\
\frac{2 b}{2 k^{2}+1}+3 \\
=\frac{1}{2 k^{2}+1}\left[3 b^{2}+2 k^{2}+2 b(2 \sqrt{2} k-1)-1\right] \\
=\frac{1}{2 k^{2}+1}(3 b+\sqrt{2} k+1)(b+\sqrt{2} k-1)
\end{array}
$$
$=0$.
Thus, $\overrightarrow{Q A} \perp \overrightarrow{Q B}$.
Clearly, points $A$, $Q$, and $B$ are distinct.
Therefore, $\angle A Q B=90^{\circ}$.
If the slope of line $l$ does not exist, then the coordinates of points $A$ and $B$ are $\left(\frac{\sqrt{2}}{3}, \pm \frac{\sqrt{17}}{3}\right)$.
It is easy to verify that $\angle A Q B=90^{\circ}$ also holds. Therefore, $\angle A Q B=90^{\circ}$.
(2) From (1), we know $\angle A Q B=90^{\circ}$.
Thus, $\triangle Q A B$ is a right triangle.
If the slope of line $Q A$ or $Q B$ does not exist, it is easy to find that $S=2 \sqrt{2}<3$.
If the slopes of lines $Q A$ and $Q B$ both exist, let $l_{Q A}: y=m(x-\sqrt{2})+1$.
Substituting into the equation of ellipse $C$ and eliminating $y$ gives $\left(2 m^{2}+1\right) x^{2}-4 m(\sqrt{2} m-1) x+$ $2(\sqrt{2} m-1)^{2}-4=0$.
Then $|Q A|$
$$
\begin{array}{l}
\Rightarrow \sqrt{m^{2}+1} \cdot \sqrt{\left[\frac{4 m(\sqrt{2} m-1)}{2 m^{2}+1}\right]^{2}-4 \cdot \frac{2(\sqrt{2} m-1)^{2}-4}{2 m^{2}+1}} \\
=\sqrt{m^{2}+1} \cdot \frac{\sqrt{8} \mid \sqrt{2} m+11}{2 m^{2}+1} .
\end{array}
$$

Since $Q B \perp Q A$, similarly,
$$
\begin{array}{l}
|Q B|=\sqrt{\left(-\frac{1}{m}\right)^{2}+1} \cdot \frac{\sqrt{8}\left|\sqrt{2}\left(-\frac{1}{m}\right)+1\right|}{2\left(-\frac{1}{m}\right)^{2}+1} \\
=\sqrt{m^{2}+1} \cdot \frac{\sqrt{8}|\sqrt{2}-m|}{m^{2}+2} .
\end{array}
$$

Thus, $S=\frac{1}{2}|Q A||Q B|$
$=4\left(m^{2}+1\right) \frac{|\sqrt{2} m+1||\sqrt{2}-m|}{\left(2 m^{2}+1\right)\left(m^{2}+2\right)}$
$=4\left(m^{2}+1\right) \frac{\left|\sqrt{2}\left(1-m^{2}\right)+m\right|}{2\left(m^{2}+1\right)^{2}+m^{2}}$
$=4 \cdot \frac{\left|\sqrt{2} \cdot \frac{1-m^{2}}{m^{2}+1}+\frac{m}{m^{2}+1}\right|}{2+\left(\frac{m}{m^{2}+1}\right)^{2}}$.
Let $\frac{1-m^{2}}{m^{2}+1}=\cos \theta, \frac{2 m}{m^{2}+1}=\sin \theta$. Then
$S=4 \cdot \frac{\left|\sqrt{2} \cos \theta+\frac{1}{2} \sin \theta\right|}{2+\frac{1}{4} \sin ^{2} \theta}$.
Notice that,
$\left|\sqrt{2} \cos \theta+\frac{1}{2} \sin \theta\right| \leqslant \sqrt{2+\frac{1}{4}}=\frac{3}{2}$,
$2+\frac{1}{4} \sin ^{2} \theta \geqslant 2$,
and the equalities cannot hold simultaneously.
Therefore, $S<4 \times \frac{\frac{3}{2}}{2}=3$.