7. Given that the ellipse $C$ passes through the point $M(1,2)$, with two foci at $(0, \pm \sqrt{6})$, and $O$ is the origin, a line $l$ parallel to $OM$ intersects the ellipse $C$ at points $A$ and $B$. Then the maximum value of the area of $\triangle OAB$ is $\qquad$.
7. 2 .

According to the problem, let $l_{A B}: y=2 x+m$.
The ellipse equation is $\frac{y^{2}}{8}+\frac{x^{2}}{2}=1$. By solving the system and eliminating $y$, we get $16 x^{2}+8 m x+2\left(m^{2}-8\right)=0$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
By Vieta's formulas, we have
$x_{1}+x_{2}=-\frac{8 m}{16}, x_{1} x_{2}=\frac{2\left(m^{2}-8\right)}{16}$.
Then $|A B|=\sqrt{5\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]}$
$$
=\frac{\sqrt{5\left(16-m^{2}\right)}}{2} \text {. }
$$

The distance from point $O$ to line $A B$ is $d=\frac{|m|}{\sqrt{5}}$, hence
$$
S_{\triangle O A B}=\frac{1}{2}|A B| d=\frac{\sqrt{m^{2}\left(16-m^{2}\right)}}{4} \leqslant 2 .
$$

When and only when $m= \pm 2 \sqrt{2}$, the area of $\triangle O A B$ reaches its maximum value of 2.