7. Ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$, a line passing through point $F(2,0)$ intersects the ellipse at points $A$ and $B$, and point $C$ is on the line $x=4$. If $\triangle A B C$ is an equilateral triangle, then the area of $\triangle A B C$ is $\qquad$
7. $\frac{72 \sqrt{3}}{25}$.

Let $l_{A B}: x=m y+2, A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, the midpoint of $A B$ be $E$. Solving the system
$$
\begin{array}{l}
\left\{\begin{array}{l}
x=m y+2, \\
x^{2}+2 y^{2}=8
\end{array}\right. \\
\Rightarrow\left(m^{2}+2\right) y^{2}+4 m y-4=0 \\
\Rightarrow|A B|=\frac{4 \sqrt{2}\left(1+m^{2}\right)}{m^{2}+2}, \\
|E C|=\sqrt{1+m^{2}}\left|4-x_{E}\right| \\
=\frac{4 \sqrt{1+m^{2}}\left(1+m^{2}\right)}{m^{2}+2} .
\end{array}
$$

Also, $\frac{|A B|}{|E C|}=\frac{2}{\sqrt{3}}$, solving for $m^{2}=\frac{1}{2}$.
Thus, $S=\frac{\sqrt{3}}{4}|A B|^{2}=\frac{72 \sqrt{3}}{25}$.