13. Given a regular tetrahedron $P-ABC$, points $P, A, B, C$ are all on a sphere with radius $\sqrt{3}$. If $PA, PB, PC$ are mutually perpendicular, then the distance from the center of the sphere to the plane $ABC$ is $\qquad$ ـ.
$=13 . \frac{\sqrt{3}}{3}$.
Let $P A=P B=P C=x$. Then $A B=\sqrt{2} x$.
Let the centroid of $\triangle A B C$ be $M$, and the center of the circumscribed sphere of the regular tetrahedron be $O, O M=y$. Then
$$
\begin{array}{l}
A M=\frac{2 \sqrt{3}}{3} \times \frac{\sqrt{2}}{2} x=\frac{\sqrt{6}}{3} x \text {. } \\
\text { Hence }\left\{\begin{array}{l}
O M^{2}+A M^{2}=A O^{2}, \\
A M^{2}+P M^{2}=A P^{2}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
y^{2}+\left(\frac{\sqrt{6}}{3} x\right)^{2}=3, \\
\left(\frac{\sqrt{6}}{3} x\right)^{2}+(\sqrt{3}-y)^{2}=x^{2}
\end{array}\right. \\
\Rightarrow y=\frac{\sqrt{3}}{3} \text { or } \sqrt{3} \text { (rejected). } \\
\end{array}
$$