11. (20 points) The ellipse $E$ with eccentricity $e=\sqrt{\frac{2}{3}}$ has its center at the origin $O$, and its foci on the $x$-axis. A line $l$ with slope $k (k \in \mathbf{R})$ passing through the point $C(-1,0)$ intersects the ellipse at points $A$ and $B$, and satisfies
$$
\overrightarrow{B A}=(\lambda+1) \overrightarrow{B C}(\lambda \geqslant 3) \text {. }
$$
(1) For a fixed $\lambda$, when the area of $\triangle O A B$ is maximized, find the equation of the ellipse $E$;
(2) If $\lambda$ varies, and $\lambda=k^{2}+1$, determine the values of the real numbers $\lambda$ and $k$ for which the length of the major axis of the ellipse $E$ is maximized. Also, find the equation of the ellipse at this time.
11. Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.

Then, from $e=\sqrt{\frac{2}{3}}$, we get $a^{2}=3 b^{2}$.
Thus, the equation of the ellipse becomes $x^{2}+3 y^{2}=3 b^{2}$.
Let $l: y=k(x+1)$, and the coordinates of the two intersection points with the ellipse be $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$.
From $\overrightarrow{C A}=\overrightarrow{B A}-\overrightarrow{B C}=\lambda \overrightarrow{B C}$, we get
$$
\left\{\begin{array}{l}
x_{1}+1=-\lambda\left(x_{2}+1\right), \\
y_{1}=-\lambda y_{2} .
\end{array}\right.
$$

By combining the equations of the line $l$ and the ellipse $E$, we get
$$
\left(3 k^{2}+1\right) x^{2}+6 k^{2} x+3\left(k^{2}-b^{2}\right)=0 \text {. }
$$

Then $\Delta=12\left[k^{2}\left(3 b^{2}-1\right)+b^{2}\right]>0$, and
$x_{1}+x_{2}=-\frac{6 k^{2}}{3 k^{2}+1}$,
$$
x_{1} x_{2}=\frac{3\left(k^{2}-b^{2}\right)}{3 k^{2}+1} \text {. }
$$

From equations (1) and (2), we get
$$
x_{2}+1=\frac{2}{(1-\lambda)\left(3 k^{2}+1\right)} \text {. }
$$
(1) When $\lambda$ is fixed,
$$
\begin{array}{l}
S_{\triangle A O B}=\frac{1}{2}\left|y_{1}-y_{2}\right||O C|=\frac{1}{2}|\lambda+1|\left|y_{2}\right| \\
=\frac{|\lambda+1|}{2}|k|\left|x_{2}+1\right|=\frac{\lambda+1}{\lambda-1} \cdot \frac{|k|}{3 k^{2}+1} \\
=\frac{\lambda+1}{\lambda-1} \cdot \frac{1}{3|k|+\frac{1}{|k|}} \leqslant \frac{\lambda+1}{\lambda-1} \cdot \frac{1}{2 \sqrt{3}},
\end{array}
$$

where the equality holds if and only if $3|k|=\frac{1}{|k|}$, i.e., $k= \pm \frac{\sqrt{3}}{3}$.
At this time, $x_{1}+x_{2}=-1$.
Combining equation (1), we get
$$
x_{1}=\frac{1}{\lambda-1}, x_{2}=-\frac{\lambda}{\lambda-1} \text {. }
$$

Substituting into equation (3), we get $3 b^{2}=\frac{\lambda^{2}+1}{(\lambda-1)^{2}}$.
At this time, the equation of the ellipse $E$ is
$$
x^{2}+3 y^{2}=\frac{\lambda^{2}+1}{(\lambda-1)^{2}}(\lambda \geqslant 3) \text {. }
$$
(2) From equations (1) and (2), we get
$$
\begin{array}{l}
x_{1}=\frac{2 \lambda}{(\lambda-1)\left(3 k^{2}+1\right)}-1, \\
x_{2}=-\frac{2}{(\lambda-1)\left(3 k^{2}+1\right)}-1 .
\end{array}
$$

Substituting into equation (3), we get
$$
3 b^{2}=\frac{4 \lambda}{(\lambda-1)^{2}\left(3 k^{2}+1\right)}+1 \text {. }
$$

From $k^{2}=\lambda-1, a^{2}=3 b^{2}$, we get
$$
\begin{array}{l}
a^{2}=\frac{4 \lambda}{(\lambda-1)^{2}(3 \lambda-2)}+1 \\
=\frac{4}{3(\lambda-1)^{2}}\left(1+\frac{2}{3 \lambda-2}\right)+1 .
\end{array}
$$

It is easy to see that when $\lambda \geqslant 3$, $a^{2}$ is a decreasing function of $\lambda$, and when $\lambda=3$, $a^{2}$ reaches its maximum value $\frac{10}{7}$.

Therefore, when $\lambda=3, k= \pm \sqrt{2}$, the length of the major axis of the ellipse $E$ reaches its maximum value. At this time, the equation of the ellipse is
$$
\frac{x^{2}}{\frac{10}{7}}+\frac{y^{2}}{\frac{10}{21}}=1 \text {. }
$$