11. In the hyperbola \( C: \frac{x^{2}}{4}-\frac{y^{2}}{5}=1 \), \( F_{1}, F_{2} \) are the left and right foci of the hyperbola \( C \), respectively, and \( P \) is a point on the hyperbola in the first quadrant. The centroid of \( \triangle P F_{1} F_{2} \) is \( G \), and the incenter is \( I \). (1) Does there exist a point \( P \) such that \( I G \parallel F_{1} F_{2} \); 
(2) Given that \( A \) is the left vertex of the hyperbola \( C \), and the line \( l \) passes through the right focus \( F_{2} \) and intersects the hyperbola \( C \) at points \( M \) and \( N \). If the slopes \( k_{1}, k_{2} \) of \( A M \) and \( A N \) satisfy \( k_{1}+k_{2}=-\frac{1}{2} \), find the equation of the line \( l \).
11. Analysis: (1) Suppose there exists a point $P$ with coordinates $\left(x_{0}, y_{0}\right)\left(y_{0}>0\right)$. Since $G$ is the centroid of $\triangle P F_{1} F_{2}$, we have $G\left(\frac{x_{0}}{3}, \frac{y_{0}}{3}\right)$. Let the inradius of $\triangle P F_{1} F_{2}$ be $r$, and $I$ be the incenter of $\triangle P F_{1} F_{2}$. Then, $S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\right| \cdot\left|y_{0}\right|=\frac{1}{2}\left(\left|P F_{1}\right|+\left|P F_{2}\right|+\left|F_{1} F_{2}\right|\right) \cdot r$. Therefore, $r=\frac{2 c y_{0}}{\left|P F_{1}\right|+\left|P F_{2}\right|+2 c}$. From $IGPF F_{1} F_{2}$, we have $\frac{2 c y_{0}}{\left|P F_{1}\right|+\left|P F_{2}\right|+2 c}=\frac{y_{0}}{3}$, which simplifies to $\left|P F_{1}\right|+\left|P F_{2}\right|=4 c$. Given $a=2, e=\frac{c}{a}$, by the focal radius formula, we have $\left|P F_{1}\right|=e x_{0}+a,\left|P F_{2}\right|=e x_{0}-a$, so $\left|P F_{1}\right|+\left|P F_{2}\right|=2 e x_{0}$. Hence, $2 e x_{0}=4 c$, which gives $x_{0}=\frac{2 c}{e}=\frac{2 \times 3}{\frac{3}{2}}=4$. Since point $P\left(x_{0}, y_{0}\right)\left(y_{0}>0\right)$ lies on the hyperbola, we have $\frac{x_{0}{ }^{2}}{4}-\frac{y_{0}{ }^{2}}{5}=1$. Solving this, we get $y_{0}=\sqrt{15}$ (the negative root is discarded). Therefore, there exists $P(4, \sqrt{15})$ such that $I G P F_{1} F_{2}$.
(2) If the slope of line $l$ does not exist, it is clear that $k_{1}+k_{2}=0$ does not satisfy the condition.
If the slope of line $l$ exists, let the equation of the line passing through $F_{2}(3,0)$ be $y=k(x-3)$, and the line intersects the ellipse at points $M\left(x_{1}, y_{1}\right)$ and $N\left(x_{2}, y_{2}\right)$. Substituting $y=k(x-3)$ into $5 x^{2}-4 y^{2}=20$, we get
$\left(5-4 k^{2}\right) x^{2}+24 k^{2} x-36 k^{2}-20=0$. By Vieta's formulas, we have: $\left\{\begin{array}{l}x_{1}+x_{2}=\frac{24 k^{2}}{4 k^{2}-5} \\ x_{1} x_{2}=\frac{36 k^{2}+20}{4 k^{2}-5}\end{array}\right.$.

Also, $k_{A M}+k_{A N}=\frac{y_{1}}{x_{1}+2}+\frac{y_{2}}{x_{2}+2}=k\left(\frac{x_{1}-3}{x_{1}+2}+\frac{x_{2}-3}{x_{2}+2}\right)=k\left[2-5\left(\frac{1}{x_{1}+2}+\frac{1}{x_{2}+2}\right)\right]$,
and $\frac{1}{x_{1}+2}+\frac{1}{x_{2}+2}=\frac{x_{1}+x_{2}+4}{x_{1} x_{2}+2\left(x_{1}+x_{2}\right)+4}=\frac{24 k^{2}+4\left(4 k^{2}-5\right)}{36 k^{2}+20+48 k^{2}+4\left(4 k^{2}-5\right)}=\frac{2 k^{2}-1}{5 k^{2}}$,
thus $k_{A M}+k_{A N}=k\left(2-5 \cdot \frac{2 k^{2}-1}{5 k^{2}}\right)=\frac{1}{k}=-\frac{1}{2}$, which gives $k=-2$. Therefore, the equation of the line $l$ is $y=-2(x-3)$.