11. (20 points) Given an ellipse with eccentricity $\frac{1}{2}$, its left focus $F_{1}$ is the intersection of the directrix of the parabola $y^{2}=4 p x(p>0)$ with the $x$-axis, and its right focus $F_{2}$ is also the focus of the parabola. The intersection point of the ellipse and the parabola above the $x$-axis is $P$. Extend $P F_{1}$ to intersect the parabola at point $Q$. $M$ is a moving point on the parabola between points $P$ and $Q$. If the side lengths of $\triangle P F_{1} F_{2}$ are exactly three consecutive positive integers, find the maximum area of $\triangle M P Q$.
11. From the problem, we know $F_{1}(-p, 0), F_{2}(p, 0)$.

Since $e=\frac{1}{2}$, the equation of the ellipse is
$$
\frac{x^{2}}{4}+\frac{y^{2}}{3}=p^{2} \text {. }
$$

From $\left\{\begin{array}{l}\frac{x^{2}}{4}+\frac{y^{2}}{3}=p^{2}, \\ y^{2}=4 p x,\end{array}\right.$, we get
$$
3 x^{2}+16 p x-12 p^{2}=0 \Rightarrow x=\frac{2}{3} p \text {. }
$$

Since point $P$ is above the $x$-axis, then $P\left(\frac{2}{3} p, \frac{2 \sqrt{6}}{3} p\right)$, and
$$
\begin{array}{l}
\left|P F_{1}\right|=\sqrt{\left(\frac{2}{3} p+p\right)^{2}+\left(\frac{2 \sqrt{6}}{3} p\right)^{2}}=\frac{7}{3} p, \\
\left|P F_{2}\right|=\frac{2}{3} p+p=\frac{5}{3} p,\left|F_{1} F_{2}\right|=2 p .
\end{array}
$$

Thus, $\frac{5}{3} p, 2 p, \frac{7}{3} p$ are consecutive positive integers.
Therefore, $p=3$.
Hence $F_{1}(-3,0), P(2,2 \sqrt{6})$,
$l_{P F_{1}}: y=\frac{2 \sqrt{6}}{5}(x+3)$.
Substituting the above into $y^{2}=12 x$, we get $Q\left(\frac{9}{2}, 3 \sqrt{6}\right)$.
Let $M\left(\frac{y_{0}^{2}}{12}, y_{0}\right)$, and $2 \sqrt{6}<y_{0}<3 \sqrt{6}$.
Since the base length $|P Q|=\frac{7}{2}$ is a constant, to maximize the area of $\triangle M P Q$, the distance from point $M$ to line $P Q$ must be maximized.
From $y^{2}=12 x \Rightarrow y=2 \sqrt{3 x} \Rightarrow y^{\prime}=\frac{\sqrt{3}}{\sqrt{x}}$.
The slope of line $P Q$ is $k_{P Q}=\frac{2 \sqrt{6}}{5}$, so the slope of the tangent line at point $M$ is
$$
\begin{array}{l}
\frac{\sqrt{3}}{\sqrt{x_{M}}}=\frac{2 \sqrt{6}}{5} \Rightarrow \frac{\sqrt{3}}{\sqrt{\frac{y_{0}^{2}}{12}}}=\frac{2 \sqrt{6}}{5} \\
\Rightarrow y_{0}=\frac{5 \sqrt{6}}{2} \Rightarrow M\left(\frac{25}{8}, \frac{5 \sqrt{6}}{2}\right) .
\end{array}
$$

The distance from point $M$ to line $P F_{1}$ is $d=\frac{\sqrt{6}}{28}$, then
$$
\left(S_{\triangle M P Q}\right)_{\max }=\frac{1}{2} \times \frac{7}{2} \times \frac{\sqrt{6}}{28}=\frac{\sqrt{6}}{16} .
$$