11. (20 points) The eccentricity \( e = \sqrt{\frac{2}{2}} \) of the ellipse \( E \) is centered at the origin \( O \), with its foci on the \( x \)-axis. A line \( l \) with slope \( k \) (where \( k \in \mathbf{R} \)) passing through the point \( C(-1,0) \) intersects the ellipse at points \( A \) and \( B \), and satisfies \( \overrightarrow{B A} = (\lambda + 1) \overrightarrow{B C} \) (\( \lambda \geqslant 3 \)).
(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 number \( \lambda \) and the slope \( k \) of the line \( l \) when 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 can be simplified to $x^{2}+3 y^{2}=b^{2}$. The equation of the line $l$ is $y=k(x+1)$, and let the coordinates of its two intersection points with the ellipse be $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. From $B A=B C+C A=(\lambda+1) B C$ we get $C A=\lambda B C$, i.e.,
$$
x_{1}+1=-\lambda\left(x_{2}+1\right), \quad y_{1}=-\lambda y_{2} .
$$

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

then $\Delta=k^{2}\left(3 b^{2}-1\right)+b^{2}>0$, and
$$
x_{1}+x_{2}=-\frac{6 k^{2}}{3 k^{2}+1}, \quad x_{1} x_{2}=\frac{3\left(k^{2}-b^{2}\right)}{3 k^{2}+1} .
$$

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

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 with equation (1) we get, $x_{1}=\frac{1}{\lambda-1}, x_{2}=-\frac{\lambda}{\lambda-1}$. Substituting into equation (2) we get $3 b^{2}=\frac{\lambda^{2}+1}{(\lambda-1)^{2}}$, thus, at this time, the equation of the ellipse $E$ is
$$
x^{2}+3 y^{2}=\frac{\lambda^{2}+1}{(\lambda-1)^{2}} \quad(\lambda \geqslant 3) \text {. }
$$
(2) From $x_{1}+1=-\lambda\left(x_{2}+1\right)$ and $x_{1}+x_{2}=-\frac{6 k^{2}}{3 k^{2}+1}$, we get
$$
x_{1}=\frac{2 \lambda}{(\lambda-1)\left(3 k^{2}+1\right)}-1, \quad x_{2}=-\frac{2}{(\lambda-1)\left(3 k^{2}+1\right)}-1,
$$

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

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

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