9. In the Cartesian coordinate system, $F_{1}, F_{2}$ are the two foci of the hyperbola $\Gamma: \frac{x^{2}}{3}-y^{2}=1$. A point $P$ on $\Gamma$ satisfies $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=1$. Find the sum of the distances from point $P$ to the two asymptotes of $\Gamma$.
Let $P(\sqrt{3} \sec \theta, \tan \theta), F_{1}(-2,0), F_{2}(2,0)$, then
$$
\begin{array}{l}
\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=(\sqrt{3} \sec \theta+2)(\sqrt{3} \sec \theta-2)+\tan ^{2} \theta=3 \sec ^{2} \theta-4+\tan ^{2} \theta=1 \\
\Rightarrow \tan ^{2} \theta=\frac{1}{2} \Rightarrow \tan \theta= \pm \frac{\sqrt{2}}{2}, \sec \theta= \pm \frac{\sqrt{6}}{2} \Rightarrow P\left( \pm \frac{3 \sqrt{2}}{2}, \pm \frac{\sqrt{2}}{2}\right) .
\end{array}
$$
The asymptote equations of $\Gamma$ are $x \pm \sqrt{3} y=0$. Taking the point $P\left(\frac{3 \sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$, the sum of the distances from point $P$ to the two asymptotes is $d_{1}+d_{2}=\frac{\left|\frac{3 \sqrt{2}}{2}+\frac{\sqrt{6}}{2}\right|}{2}+\frac{\left|\frac{3 \sqrt{2}}{2}-\frac{\sqrt{6}}{2}\right|}{2}=\frac{3 \sqrt{2}}{2}$.
By symmetry, the sum of the distances from the other points $P$ to the two asymptotes of $\Gamma$ is also $\frac{3 \sqrt{2}}{2}$.