9. (14 points) Given that $E$ is a point on the edge $BC$ of a regular tetrahedron $ABCD$ with edge length 2. The inradii of the tetrahedra $ABDE$ and $ACDE$ are $r_{1}$ and $r_{2}$, respectively. If $r_{1}^{-1}+r_{2}^{-1}=4 \sqrt{6}+4 \sqrt{3}$, then there are two points $E$ that satisfy the condition, denoted as $E_{1}$ and $E_{2}$. Find the length of $E_{1} E_{2}$.
Let $F$ be the midpoint of $BC$, and $EF = x$. By the Pythagorean theorem, we have
$$
\begin{array}{l}
V_{\text {tetrahedron } 1 C D E}=\frac{\sqrt{2}}{3}(1+x), \\
S_{\triangle A B E}=S_{\triangle D B E}=\frac{\sqrt{3}}{2}(1-x), \\
S_{\triangle M C E}=S_{\triangle D C E}=\frac{\sqrt{3}}{2}(1+x),
\end{array}
$$

Then,
$$
r_{1}=\frac{\sqrt{2}(1-x)}{\sqrt{3}+\sqrt{3}(1-x)+\sqrt{x^{2}+2}},
$$
$$
r_{2}=\frac{\sqrt{2}(1+x)}{\sqrt{3}+\sqrt{3}(1+x)+\sqrt{x^{2}+2}}.
$$

Thus,
$$
r_{1}^{-1}+r_{2}^{-1}
=\sqrt{6}+\left(\sqrt{3}+\sqrt{x^{2}+2}\right) \frac{\sqrt{2}}{1-x^{2}}=4 \sqrt{6}+4 \sqrt{3}.
$$

Therefore,
$$
\frac{\sqrt{3}+\sqrt{x^{2}+2}}{1-x^{2}}=3 \sqrt{3}+2 \sqrt{6}.
$$
Solving for $x$, we get $x = \pm \sqrt{\frac{2}{3}}$.
Hence, $E_{1} E_{2}=2 \sqrt{\frac{2}{3}}=\frac{2 \sqrt{6}}{3}$.