12. Given $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, respectively. 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 $C$. A line $l$ passing through point $F_{2}$ intersects the ellipse $C$ at points $D$ and $E$. Denote the slopes of lines $A D$ and $A E$ as $k_{1}$ and $k_{2}$, respectively. If $k_{1}+k_{2}=-\frac{1}{2}$, find the equation of the line $l$.
12. Let $F_{1}(-c, 0), F_{2}(c, 0)$.

From the given information, we have $F_{1} H \perp P F_{2}$.

Then
$$
\begin{array}{l}
k_{F_{1} H} k_{P F_{2}}=\frac{-\frac{5}{3}}{\frac{2 \sqrt{6}}{3}+c} \cdot \frac{1}{\frac{2 \sqrt{6}}{3}-c}=-1 \\
\Rightarrow \frac{24}{9}-c^{2}=\frac{5}{3} \\
\Rightarrow c^{2}=1 .
\end{array}
$$

Since point $P\left(\frac{2 \sqrt{6}}{3}, 1\right)$ lies on the ellipse $C$,
$$
\frac{24}{9 a^{2}}+\frac{1}{b^{2}}=1 \text {. }
$$

Combining with $a^{2}-b^{2}=c^{2}=1$, we solve to get $a^{2}=4, b^{2}=3$.
Thus, 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, by symmetry, we have $k_{1}+k_{2}=0$, which does not meet the requirement.
If the slope of $l$ exists, let it be $k$, then the equation of $l$ is
$$
y=k(x-1) \text {. }
$$

Substituting the above into the equation of the ellipse, we get
$$
\left(4 k^{2}+3\right) x^{2}-8 k^{2} x+4 k^{2}-12=0 \text {. }
$$

Let $D\left(x_{1}, y_{1}\right), E\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}+x_{2}=\frac{8 k^{2}}{4 k^{2}+3}, x_{1} x_{2}=\frac{4 k^{2}-12}{4 k^{2}+3} \text {. }
$$

Thus, $k_{1}+k_{2}=\frac{y_{1}}{x_{1}+2}+\frac{y_{2}}{x_{2}+2}$
$$
\begin{array}{l}
=\frac{k\left(x_{1}-1\right)}{x_{1}+2}+\frac{k\left(x_{2}-1\right)}{x_{2}+2} \\
=k\left(2-\frac{3\left(x_{1}+x_{2}+4\right)}{x_{1} x_{2}+2\left(x_{1}+x_{2}\right)+4}\right) \\
=k\left(2-\frac{3\left(\frac{8 k^{2}}{4 k^{2}+3}+4\right)}{\frac{4 k^{2}-12}{4 k^{2}+3}+2 \cdot \frac{8 k^{2}}{4 k^{2}+3}+4}\right) \\
=k\left(2-\frac{2 k^{2}+1}{k^{2}}\right)=-\frac{1}{k}=-\frac{1}{2} .
\end{array}
$$

Therefore, $k=2$, and the line $l: y=2(x-1)$.