9. (This question is worth 16 points) In the ellipse $\Gamma$, $F$ is a focus, and $A, B$ are two vertices. If $|F A|=3,|F B|=2$, find all possible values of $|A B|$.

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Solution: Let's assume the standard equation of the ellipse $\Gamma$ in the Cartesian coordinate system is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and denote $c=\sqrt{a^{2}-b^{2}}$. By symmetry, we can assume $F$ is the right focus of $\Gamma$.

It is easy to know that the distance from $F$ to the left vertex of $\Gamma$ is $a+c$, to the right vertex is $a-c$, and to the top and bottom vertices is $a$. We discuss the following cases:
(1) $A, B$ are the left and right vertices, respectively. In this case, $a+c=3, a-c=2$, so $|A B|=2 a=5$ (correspondingly, $b^{2}=(a+c)(a-c)=6$, and the equation of $\Gamma$ is $\left.\frac{4 x^{2}}{25}+\frac{y^{2}}{6}=1\right)$.
(2) $A$ is the left vertex, and $B$ is the top or bottom vertex. In this case, $a+c=3, a=2$, so $c=1$, and thus $b^{2}=a^{2}-c^{2}=3$, so $|A B|=\sqrt{a^{2}+b^{2}}=\sqrt{7}$ (correspondingly, the equation of $\Gamma$ is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ ).
(3) $A$ is the top or bottom vertex, and $B$ is the right vertex. In this case, $a=3, a-c=2$, so $c=1$, and thus $b^{2}=a^{2}-c^{2}=8$, so $|A B|=\sqrt{a^{2}+b^{2}}=\sqrt{17}$ (correspondingly, the equation of $\Gamma$ is $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$ ).
In summary, all possible values of $|A B|$ are $5, \sqrt{7}, \sqrt{17}$. 12 points. 16 points.