11. (20 points) In the hyperbola \( C: \frac{x^{2}}{4}-\frac{y^{2}}{5}=1 \), \( F_{1} \) and \( F_{2} \) are the left and right foci of the hyperbola \( C \), respectively. \( P \) is a point on the hyperbola and in the first quadrant, \( \triangle P F_{1} F_{2} \) has a centroid \( G \) and an incenter \( I \).
(1) Does there exist a point \( P \) such that \( IG \parallel F_{1} F_{2} \)?
(2) Let \( A \) be the left vertex of the hyperbola \( C \), and let the line \( l \) pass through the right focus \( F_{2} \), intersecting the hyperbola \( C \) at points \( M \) and \( N \). If the slopes \( k_{1} \) and \( k_{2} \) of \( AM \) and \( AN \) satisfy \( k_{1} + k_{2} = -\frac{1}{2} \), find the equation of the line \( l \).
11. (1) Suppose there exists a point $P\left(x_{0}, y_{0}\right)\left(y_{0}>0\right)$.

Since $G$ is the centroid of $\triangle P F_{1} F_{2}$, we know $G\left(\frac{x_{0}}{3}, \frac{y_{0}}{3}\right)$.
Let the inradius of $\triangle P F_{1} F_{2}$ be $r$.
Then $S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\right|\left|y_{0}\right|$
$=\frac{1}{2}\left(\left|P F_{1}\right|+\left|P F_{2}\right|+\left|F_{1} F_{2}\right|\right) r$
$\Rightarrow \frac{1}{2} \times 2 c\left|y_{0}\right|=\frac{1}{2}\left(\left|P F_{1}\right|+\left|P F_{2}\right|+2 c\right) r$
$\Rightarrow r=\frac{2 c y_{0}}{\left|P F_{1}\right|+\left|P F_{2}\right|+2 c}$.
Since $I G / / F_{1} F_{2}$
$\Rightarrow \frac{2 c y_{0}}{\left|P F_{1}\right|+\left|P F_{2}\right|+2 c}=\frac{y_{0}}{3}$
$\Rightarrow\left|P F_{1}\right|+\left|P F_{2}\right|=4 c$.
By the focal radius formula, we know
$\left|P F_{1}\right|=e x_{0}+a,\left|P F_{2}\right|=e x_{0}-a$
$\Rightarrow\left|P F_{1}\right|+\left|P F_{2}\right|=2 e x_{0}$
$\Rightarrow 2 e x_{0}=4 c$
$\Rightarrow x_{0}=\frac{2 c}{e}=2 a=4$.
Since point $P$ is on the hyperbola, we have
$\frac{x_{0}^{2}}{4}-\frac{y_{0}^{2}}{5}=1 \Rightarrow y_{0}=\sqrt{15}$ (negative value discarded).
Thus, there exists a point $P(4, \sqrt{15})$, such that $I G / / F_{1} F_{2}$.
(2) If the slope of line $l$ does not exist, it is clear that $k_{1}+k_{2}=0$ does not meet the condition.

If the slope of line $l$ exists, let the equation of the line passing through point $F_{2}(3,0)$ be $y=k(x-3)$, and the line intersects the ellipse at points $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$.
Substituting $y=k(x-3)$ into $5 x^{2}-4 y^{2}=20$, we get
$$
\left(5-4 k^{2}\right) x^{2}+24 k^{2} x-36 k^{2}-20=0 \text {. }
$$

By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=\frac{24 k^{2}}{4 k^{2}-5}, \\
x_{1} x_{2}=\frac{36 k^{2}+20}{4 k^{2}-5} .
\end{array}\right.
$$

Then $k_{A M}+k_{A N}=\frac{y_{1}}{x_{1}+2}+\frac{y_{2}}{x_{2}+2}$
$$
\begin{array}{l}
=k\left(\frac{x_{1}-3}{x_{1}+2}+\frac{x_{2}-3}{x_{2}+2}\right) \\
=k\left[2-5\left(\frac{1}{x_{1}+2}+\frac{1}{x_{2}+2}\right)\right] \\
=k\left(2-5 \times \frac{2 k^{2}-1}{5 k^{2}}\right)=\frac{1}{k}=-\frac{1}{2} \\
\Rightarrow k=-2 .
\end{array}
$$

Thus, the equation of the line $l$ is $y=-2(x-3)$.