12. Given $F_{1}, F_{2}$ are the left and right foci of the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, point $P\left(\frac{2 \sqrt{6}}{3}, 1\right)$ lies on the ellipse $C$, and the orthocenter of $\triangle F_{1} P F_{2}$ is $H\left(\frac{2 \sqrt{6}}{3},-\frac{5}{3}\right)$.
(1) Find the equation of the ellipse $C$;
(2) Let $A$ be the left vertex of the ellipse $C$, and let the line $l$ passing through point $F_{2}$ intersect the ellipse $C$ at points $D, E$. Denote the slopes of the lines $A D, A E$ as $k_{1}, k_{2}$, respectively. If $k_{1}+k_{2}=-\frac{1}{2}$, find the equation of the line $l$.
(1) Let $F_{1}(-c, 0), F_{2}(c, 0)$, and given that the orthocenter of $\triangle F_{1} P F_{2}$ is $H\left(\frac{2 \sqrt{6}}{3},-\frac{5}{3}\right)$, we have $F_{1} H \perp P F_{2}$, so
$$
k_{F_{1} H} \cdot k_{P F_{2}}=\frac{-\frac{5}{3}}{\frac{2 \sqrt{6}}{3}+c} \cdot \frac{1}{\frac{2 \sqrt{6}}{3}-c}=-1, \frac{24}{9}-c^{2}=\frac{5}{3}
$$

Solving for $c^{2}$, we get $c^{2}=1$.
Since point $P\left(\frac{2 \sqrt{6}}{3}, 1\right)$ lies on the ellipse $C$, we have $\frac{24}{9 a^{2}}+\frac{1}{b^{2}}=1$, and combining with $a^{2}-b^{2}=c^{2}=1$, we solve to get $a^{2}=4, b^{2}=3$. Therefore, the equation of the ellipse $C$ is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$;
(2) From (1), we know $A(-2,0), F_{2}(1,0)$.
If the slope of $l$ does not exist, then by symmetry, $k_{1}+k_{2}=0$, which does not meet the requirement;
If the slope of $l$ exists, let it be $k$, then the equation of $l$ is $y=k(x-1)$.
From $\left\{\begin{array}{l}y=k(x-1) \\ \frac{x^{2}}{4}+\frac{y^{2}}{3}=1\end{array}\right.$, we get $\left(4 k^{2}+3\right) x^{2}-8 k^{2} x+4 k^{2}-12=0$.
Let $D\left(x_{1}, y_{1}\right), E\left(x_{2}, y_{2}\right)$, then $x_{1}+x_{2}=\frac{8 k^{2}}{4 k^{2}+3}, x_{1} x_{2}=\frac{4 k^{2}-12}{4 k^{2}+3}$.
Therefore,
$$
\begin{aligned}
k_{1}+k_{2} & =\frac{y_{1}}{x_{1}+2}+\frac{y_{2}}{x_{2}+2}=\frac{k\left(x_{1}-1\right)}{x_{1}+2}+\frac{k\left(x_{2}-1\right)}{x_{2}+2}=k\left(1-\frac{3}{x_{1}+2}+1-\frac{3}{x_{2}+2}\right) \\
& =k \cdot\left[2-\frac{3\left(x_{1}+x_{2}+4\right)}{\left(x_{1}+2\right)\left(x_{2}+2\right)}\right]=k \cdot\left[2-\frac{3\left(x_{1}+x_{2}+4\right)}{x_{1} x_{2}+2\left(x_{1}+x_{2}\right)+4}\right]=k \cdot\left[2-\frac{3\left(\frac{8 k^{2}}{4 k^{2}+3}+4\right)}{\frac{4 k^{2}-12}{4 k^{2}+3}+2 \times \frac{8 k^{2}}{4 k^{2}+3}+4}\right] \\
& =k \cdot\left[2-\frac{3\left(8 k^{2}+16 k^{2}+12\right)}{4 k^{2}-12+16 k^{2}+16 k^{2}+12}\right]=k \cdot\left(2-\frac{2 k^{2}+1}{k^{2}}\right)=-\frac{1}{k}
\end{aligned}
$$

Since $k_{1}+k_{2}=-\frac{1}{2}$, we have $k=2$, and the equation of the line $l$ is $y=2(x-1)$.
(20 points)