9. Given a tetrahedron $P-ABC$ with the base being an equilateral triangle of side length $4 \sqrt{3}$, $PA=3, PB=4, PC=5$. If $O$ is the center of $\triangle ABC$, then the length of $PO$ is $\qquad$ .
9. $\sqrt{6}$.

Solution 1 As shown in Figure 6, connect $A O$ and extend it to intersect $B C$ at point $M$, then $M$ is the midpoint of side $B C$. Connect $P M$.
In $\triangle P B C$, by the median length formula, we get $P M=\frac{5}{\sqrt{2}}$.
In $\triangle P A M$, by the cosine rule, we get
$$
\cos \angle P A M=\frac{41}{24 \sqrt{6}} \text {. }
$$

In $\triangle P O A$, by the cosine rule, we get
$$
P O=\sqrt{P A^{2}+A O^{2}-2 P A \cdot A O \cos \angle P A M}=\sqrt{6} \text {. }
$$

Solution 2 Let $\overrightarrow{P A}=a, \overrightarrow{P B}=b, \overrightarrow{P C}=c$.
Then, by $\overrightarrow{P B}-\overrightarrow{P A}=\overrightarrow{A B}$, we get
$$
(b-a)^{2}=\overrightarrow{A B}^{2}=32 \text {. }
$$

Thus, $a^{2}+b^{2}-2 a \cdot b=32 \Rightarrow 2 a \cdot b=-7$.
Similarly, $2 b \cdot c=9,2 c \cdot a=2$.
Since $O$ is the center of the equilateral $\triangle A B C$, we have,
$$
\begin{array}{l}
\overrightarrow{P O}=\frac{1}{3}(a+b+c) \text {. } \\
\text { Therefore, } \overrightarrow{P O}^{2}=\frac{1}{9}(a+b+c)^{2} \\
=\frac{1}{9}\left(a^{2}+b^{2}+c^{2}+2 a \cdot b+2 b \cdot c+2 c \cdot a\right) \\
=6 \text {. } \\
\end{array}
$$

Therefore, the length of $P O$ is $\sqrt{6}$.