Let the equation of $\Gamma$ be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
From $b^{2}=\left(1-e^{2}\right) a^{2}=\frac{1}{3} a^{2}$, we know that the equation of $\Gamma$ can be transformed into $x^{2}+3 y^{2}=a^{2}$.
From $C A=2 B C$, we know that $l$ is not parallel to the coordinate axes. We can set $l: y=$ $k(x+1)$, which can be transformed into $x=\frac{1}{k} y-1$, and substituting into the equation of $\Gamma$ and simplifying, we get
$$\left(\frac{1}{k^{2}}+3\right) y^{2}-\frac{2}{k} y+1-a^{2}=0 .$$

To ensure that $A$ and $B$ exist, we should have
$$\Delta=\frac{4}{k^{2}}-4\left(\frac{1}{k^{2}}+3\right)\left(1-a^{2}\right)>0 \text {. }$$

Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, then we have
$$y_{1}+y_{2}=\frac{2 k}{1+3 k^{2}} \text {. }$$

From $C A=2 B C$, we know that $y_{1}=-2 y_{2}$. Substituting into the above equation, we get $\boldsymbol{y}_{2}=\frac{-2 k}{1+3 k^{2}}$. Thus,
$$\begin{array}{l} S_{\triangle O A B}=\frac{1}{2}|O C| \cdot\left|y_{1}-y_{2}\right|=\frac{3}{2}\left|y_{2}\right| \\ =\frac{3|k|}{1+3 k^{2}} \leqslant \frac{3|k|}{2 \sqrt{3}|k|}=\frac{\sqrt{3}}{2}, \end{array}$$

where the equality holds when $k= \pm \frac{\sqrt{3}}{3}$.
From $y_{2}=\frac{-2 k}{1+3 k^{2}}$, we can get $y_{2}=\mp \frac{\sqrt{3}}{3}$.
Substituting the two sets of $k$ and $y$ values into equation (1), we can solve for $a^{2}=5$.
Therefore, the maximum value of the area of $\triangle O A B$ is $\frac{\sqrt{3}}{2}$, and the corresponding equation of $\Gamma$ is $x^{2}+3 y^{2}=5$.
Editor's note: In this solution, it is assumed that point $C$ is inside the ellipse. When point $C$ is outside the ellipse, $y_{1}=2 y_{2}, y_{2}=\frac{2 k}{3\left(1+3 k^{2}\right)}, S_{\triangle O A B}=$ $\frac{|k|}{3\left(1+3 k^{2}\right)} \leqslant \frac{\sqrt{3}}{18}<\frac{\sqrt{3}}{2}$.