9. Given $\triangle A B C$ and a point $D$ on its side $B C$ such that: $A B=2 B D, A C=3 C D$, and an ellipse $\Gamma$ can be drawn with $A, D$ as foci, passing through points $B, C$. Find the eccentricity of $\Gamma$.

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Given $B D=x, D C=y$, according to the problem, $3 x=4 y$. Taking $B C=7$, then $A B=8, A C=9 \Rightarrow \cos \angle B A C=\frac{8^{2}+9^{2}-7^{2}}{2 \cdot 8 \cdot 9}=\frac{2}{3}$. From $\overrightarrow{B D}=\frac{4}{7} \overrightarrow{B C} \Rightarrow \overrightarrow{A D}=\frac{4}{7} \overrightarrow{A C}+\frac{3}{7} \overrightarrow{A B}$
$\Rightarrow A D^{2}=\frac{16 \cdot 81+9 \cdot 64+24 \cdot 48}{49}=\frac{2^{4} \cdot 3^{2} \cdot 21}{49}$
$\Rightarrow A D=\frac{12 \sqrt{21}}{7}$. Thus, for the ellipse $\Gamma$, $2 a=12,2 c=\frac{12 \sqrt{21}}{7}$,
so the eccentricity of the ellipse $\Gamma$ is $e=\frac{c}{a}=\frac{\sqrt{21}}{7}$.