11 Given the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{2}=1$, a moving line $l$ passes through the point $P\left(\frac{\sqrt{2}}{3},-\frac{1}{3}\right)$ but does not pass through the point $Q(\sqrt{2}, 1)$, and 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 $-\frac{1}{3}=\frac{\sqrt{2}}{3} k+b$, hence $b=-\frac{1}{3}(\sqrt{2} k+1)$.
By combining the equation of line $l$ and the equation of ellipse $C$, and eliminating $y$, we get
$$
\left(2 k^{2}+1\right) x^{2}+4 k b x+2 b^{2}-4=0 .
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then
$$
\begin{array}{c}
x_{1}+x_{2}=-\frac{4 k b}{2 k^{2}+1}, \\
x_{1} x_{2}=\frac{2 b^{2}-4}{2 k^{2}+1}, \\
y_{1}+y_{2}=k\left(x_{1}+x_{2}\right)+2 b \\
=-\frac{4 k^{2} b}{2 k^{2}+1}+2 b \\
=\frac{2 b}{2 k^{2}+1}, \\
y_{1} y_{2}=\left(k x_{1}+b\right)\left(k x_{2}+b\right) \\
=k^{2} x_{1} x_{2}+k b\left(x_{1}+x_{2}\right)+b^{2} \\
=k^{2} \cdot \frac{2 b^{2}-4}{2 k^{2}+1}+k b \cdot\left(-\frac{4 k b}{2 k^{2}+1}\right)+b^{2} \\
=\frac{b^{2}-4 k^{2}}{2 k^{2}+1} .
\end{array}
$$

Since
$$
\begin{array}{l}
\overrightarrow{Q A}=\left(x_{1}-\sqrt{2}, y_{1}-1\right), \\
\overrightarrow{Q B}=\left(x_{2}-\sqrt{2}, y_{2}-1\right),
\end{array}
$$

we have
$$
\begin{aligned}
\overrightarrow{Q A} \cdot \overrightarrow{Q B}= & \left(x_{1}-\sqrt{2}, y_{1}-1\right) \cdot\left(x_{2}-\sqrt{2}, y_{2}-1\right) \\
= & \left(x_{1}-\sqrt{2}\right)\left(x_{2}-\sqrt{2}\right)+\left(y_{1}-1\right)\left(y_{2}-1\right) \\
= & x_{1} x_{2}-\sqrt{2}\left(x_{1}+x_{2}\right)+2+y_{1} y_{2}-\left(y_{1}+y_{2}\right)+1 \\
= & \frac{2 b^{2}-4}{2 k^{2}+1}-\sqrt{2} \cdot\left(-\frac{4 k b}{2 k^{2}+1}\right)+2+\frac{b^{2}-4 k^{2}}{2 k^{2}+1}- \\
& \frac{2 b}{2 k^{2}+1}+1 \\
= & \frac{1}{2 k^{2}+1}\left[3 b^{2}+2 k^{2}+2 b(2 \sqrt{2} k-1)-1\right] \\
= & \frac{1}{2 k^{2}+1}\left[\frac{1}{3}(\sqrt{2} k+1)^{2}+2 k^{2}-\right. \\
& \left.\frac{2}{3}(\sqrt{2} k+1)(2 \sqrt{2} k-1)-1\right] \\
= & 0,
\end{aligned}
$$

Therefore, $\overrightarrow{Q A} \perp \overrightarrow{Q B}$, and clearly points $A$, $Q$, and $B$ are distinct, so $\angle A Q B=90^{\circ}$.
If the slope of line $l$ does not exist, then the coordinates of points $A$ and $B$ are $\left(\frac{\sqrt{2}}{3}\right.$, $\left.\pm \frac{\sqrt{17}}{3}\right)$, and it is easy to verify that $\angle A Q B=90^{\circ}$ also holds.
Thus, $\angle A Q B=90^{\circ}$.
(2) From (1), we know that $\angle A Q B=90^{\circ}$, so $\triangle Q A B$ is a right triangle.
If the slope of line $Q A$ or $Q B$ does not exist, it is easy to find that the area of $\triangle Q A B$ is $S=2 \sqrt{2}<3$.

If the slopes of lines $Q A$ and $Q B$ both exist, let the equation of line $Q A$ be $y=m(x-\sqrt{2})+1$, and substitute it into the equation of ellipse $C$, eliminating $y$, we get
$$
\left(2 m^{2}+1\right) x^{2}-4 m(\sqrt{2} m-1) x+2(\sqrt{2} m-1)^{2}-4=0,
$$

then
$$
\begin{aligned}
|Q A| & =\sqrt{m^{2}+1} \cdot \sqrt{\left(\frac{4 m(\sqrt{2} m-1)}{2 m^{2}+1}\right)^{2}-4 \cdot \frac{2(\sqrt{2} m-1)^{2}-4}{2 m^{2}+1}} \\
& =\sqrt{m^{2}+1} \cdot \frac{\sqrt{8} \cdot|\sqrt{2} m+1|}{2 m^{2}+1} .
\end{aligned}
$$

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

Thus, the area of $\triangle Q A B$ is
$$
\begin{aligned}
S & =\frac{1}{2}|Q A| \cdot|Q B| \\
& =\frac{1}{2} \cdot \sqrt{m^{2}+1} \cdot \frac{\sqrt{8} \cdot|\sqrt{2} m+1|}{2 m^{2}+1} \cdot \sqrt{m^{2}+1} \cdot \frac{\sqrt{8} \cdot|\sqrt{2}-m|}{m^{2}+2} \\
& =4\left(m^{2}+1\right) \cdot \frac{|\sqrt{2} m+1| \cdot|\sqrt{2}-m|}{\left(2 m^{2}+1\right)\left(m^{2}+2\right)} \\
& =4\left(m^{2}+1\right) \cdot \frac{\left|\sqrt{2}\left(1-m^{2}\right)+m\right|}{2\left(m^{2}+1\right)^{2}+m^{2}} \\
= & 4 \cdot \frac{\left|\sqrt{2} \cdot \frac{1-m^{2}}{m^{2}+1}+\frac{m}{m^{2}+1}\right|}{2+\left(\frac{m}{m^{2}+1}\right)^{2}} . \\
& \text { Let } \frac{1-m^{2}}{m^{2}+1}=\cos \theta, \frac{2 m}{m^{2}+1}=\sin \theta, \text { then } \\
& S=4 \cdot \frac{\left|\sqrt{2} \cos \theta+\frac{1}{2} \sin \theta\right|}{2+\frac{1}{4} \sin ^{2} \theta} .
\end{aligned}
$$

Notice that
$$
\begin{array}{l}
\left|\sqrt{2} \cos \theta+\frac{1}{2} \sin \theta\right|=\sqrt{2+\frac{1}{4}} \cdot|\sin (\theta+\varphi)| \\
\leqslant \sqrt{2+\frac{1}{4}}=\frac{3}{2}, \\
2+\frac{1}{4} \sin ^{2} \theta \geqslant 2,
\end{array}
$$

and the equalities cannot hold simultaneously, so
$$
S<4 \cdot \frac{\frac{3}{2}}{2}=3 \text {. }
$$