11. (20 points) Given the ellipse $\Gamma: \frac{x^{2}}{4}+y^{2}=1$, point $P(1,0)$, lines $l_{1}: x=-2, l_{2}: x=2$, $AB$ and $CD$ are two chords of the ellipse $\Gamma$ passing through point $P$, and the line $l_{CD}$: $x=1$, lines $AC$ and $BD$ intersect $l_{2}$ and $l_{1}$ at points $E$ and $F$ respectively. Find the ratio of the y-coordinates of points $E$ and $F$.
11. Let's assume point \( C\left(1, \frac{\sqrt{3}}{2}\right), D\left(1,-\frac{\sqrt{3}}{2}\right) \)
\[
\begin{array}{l}
l_{A B}: t y=x-1, A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), \\
E(2, e), F(-2, f) .
\end{array}
\]

By combining the equation of line \( A B \) with the equation of the ellipse \( \Gamma \), we get
\[
\begin{array}{l}
(t y+1)^{2}+4 y^{2}-4=0 \\
\Rightarrow\left(t^{2}+4\right) y^{2}+2 t y-3=0 .
\end{array}
\]

By Vieta's formulas, we have
\[
y_{1}+y_{2}=\frac{-2 t}{t^{2}+4}, y_{1} y_{2}=\frac{-3}{t^{2}+4} \text {. }
\]

Since points \( A, C, E \) are collinear,
\[
\begin{aligned}
\Rightarrow & \frac{x_{1}-1}{2-1}=\frac{y_{1}-\frac{\sqrt{3}}{2}}{e-\frac{\sqrt{3}}{2}} \\
\Rightarrow e & =\frac{y_{1}-\frac{\sqrt{3}}{2}}{x_{1}-1}+\frac{\sqrt{3}}{2} \\
& =\frac{y_{1}-\frac{\sqrt{3}}{2}}{t y_{1}}+\frac{\sqrt{3}}{2} \\
& =\frac{1}{t}+\frac{\sqrt{3}}{2}\left(1-\frac{1}{t y_{1}}\right) .
\end{aligned}
\]

Since points \( B, D, F \) are collinear,
\[
\begin{aligned}
\Rightarrow & \frac{x_{2}-1}{-2-1}=\frac{y_{2}-\frac{-\sqrt{3}}{2}}{f-\frac{-\sqrt{3}}{2}} \\
\Rightarrow f & =\frac{-3\left(y_{2}+\frac{\sqrt{3}}{2}\right)}{x_{2}-1}-\frac{\sqrt{3}}{2} \\
& =\frac{-3\left(y_{2}+\frac{\sqrt{3}}{2}\right)}{t y_{2}}-\frac{\sqrt{3}}{2} \\
& =\frac{-3}{t}-\frac{\sqrt{3}}{2}\left(1+\frac{3}{t y_{2}}\right) .
\end{aligned}
\]

Then \( 3 e+f \)
\[
\begin{array}{l}
=3\left(\frac{1}{t}+\frac{\sqrt{3}}{2} \cdot \frac{t y_{1}-1}{t y_{1}}\right)+\frac{-3}{t}-\frac{\sqrt{3}}{2} \cdot \frac{t y_{2}+3}{t y_{2}} \\
=\frac{\sqrt{3}}{2}\left(3-\frac{3}{t y_{1}}-1-\frac{3}{t y_{2}}\right) .
\end{array}
\]

Since \(\frac{1}{t y_{1}}+\frac{1}{t y_{2}}=\frac{y_{1}+y_{2}}{t y_{1} y_{2}}=\frac{\frac{-2 t}{t^{2}+4}}{t \cdot \frac{-3}{t^{2}+4}}=\frac{2}{3}\), we have
\[
3 e+f=0 \text {. }
\]

Therefore, the required value is \( -\frac{1}{3} \).