11. (20 points) Through the point \( P\left(1, \frac{1}{2}\right) \) inside the ellipse \(\frac{x^{2}}{6}+\frac{y^{2}}{3}=1\), draw a line that does not pass through the origin, intersecting the ellipse at points \( A \) and \( B \). Find the maximum value of \( S_{\triangle O A B} \).
11. Let $A(\sqrt{6} \cos \alpha, \sqrt{3} \sin \alpha)$, $B(\sqrt{6} \cos \beta, \sqrt{3} \sin \beta)$.
Assume $0 \leqslant \beta < \alpha < 2 \pi$.
When $A B \perp x$-axis, we get
$S_{\triangle O A B}=\frac{\sqrt{10}}{2}$.
When $A B$ is not perpendicular to the $x$-axis,
$$
\begin{aligned}
k_{A B} & =\frac{\sqrt{3} \sin \beta - \sqrt{3} \sin \alpha}{\sqrt{6} \cos \beta - \sqrt{6} \cos \alpha} = \frac{\sqrt{3} \sin \alpha - \frac{1}{2}}{\sqrt{6} \cos \alpha - 1} \\
\Rightarrow & \sqrt{6}(\sin \alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta) \\
& = \left(\sin \alpha - \frac{\sqrt{2}}{2} \cos \alpha\right) - \left(\sin \beta - \frac{\sqrt{2}}{2} \cos \beta\right) \\
\Rightarrow & \sin (\alpha - \beta) = \frac{1}{2} \sin (\alpha - \varphi) - \frac{1}{2} \sin (\beta - \varphi) \\
& \left(\tan \varphi = -\frac{\sqrt{2}}{2}\right) \\
\Rightarrow & 2 \sin \frac{\alpha - \beta}{2} \cdot \cos \frac{\alpha - \beta}{2} \\
= & \sin \frac{\alpha - \beta}{2} \cdot \cos \frac{\alpha + \beta - 2 \varphi}{2} \\
\Rightarrow & \cos \frac{\alpha - \beta}{2} = \frac{1}{2} \cos \frac{\alpha + \beta - 2 \varphi}{2} .
\end{aligned}
$$

From $-\varphi \leqslant \frac{\alpha + \beta - 2 \varphi}{2} \leqslant 2 \pi - \varphi$
$$
\begin{array}{l}
\Rightarrow -1 \leqslant \cos \frac{\alpha + \beta - 2 \varphi}{2} \leqslant 1 \\
\Rightarrow -\frac{1}{2} \leqslant \cos \frac{\alpha - \beta}{2} \leqslant \frac{1}{2} \\
\Rightarrow \frac{\pi}{3} \leqslant \frac{\alpha - \beta}{2} \leqslant \frac{2 \pi}{3} \\
\Rightarrow \frac{2 \pi}{3} \leqslant \alpha - \beta \leqslant \frac{4 \pi}{3} \\
\Rightarrow |\sin (\alpha - \beta)| \leqslant \frac{\sqrt{3}}{2} .
\end{array}
$$

The equation of the line $A B$ is
$$
y = \frac{\sin \alpha - \sin \beta}{\sqrt{2}(\cos \alpha - \cos \beta)}(x - 1) + \frac{1}{2},
$$

When $y = 0$,
$$
\begin{aligned}
x & = 1 - \frac{\cos \alpha - \cos \beta}{\sqrt{2}(\sin \alpha - \sin \beta)} \\
& = \frac{\sqrt{6} \sin (\alpha - \beta)}{\sin \alpha - \sin \beta},
\end{aligned}
$$

Thus, the point where line $A B$ intersects the $x$-axis is
$$
\begin{array}{l}
M\left(\frac{\sqrt{6} \sin (\alpha - \beta)}{\sin \alpha - \sin \beta}, 0\right) . \\
\text{Then } S_{\triangle O A B} = \frac{1}{2} \left|x_{M}\right| \left|y_{A} - y_{B}\right| \\
= \frac{1}{2} \left|\frac{\sqrt{6} \sin (\alpha - \beta)}{\sin \alpha - \sin \beta}\right| |\sqrt{3}(\sin \alpha - \sin \beta)| \\
= \frac{3 \sqrt{2}}{2} |\sin (\alpha - \beta)| \\
\leqslant \frac{3 \sqrt{2}}{2} \times \frac{\sqrt{3}}{2} = \frac{3 \sqrt{6}}{4} .
\end{array}
$$

When $\alpha - \beta = \frac{2 \pi}{3}$ or $\alpha - \beta = \frac{4 \pi}{3}$, the equality holds.
Upon verification, when $\alpha - \beta = \frac{2 \pi}{3}$ or $\alpha - \beta = \frac{4 \pi}{3}$,
$$
\cos \frac{\alpha - \beta}{2} = \frac{1}{2} \cos \frac{\alpha + \beta - \varphi}{2} \text{.}
$$

Thus, $\left(S_{\triangle O A B}\right)_{\max} = \frac{3 \sqrt{6}}{4}$.