10 From a point $O$ on a plane, two rays $O A$ and $O B$ are drawn. $P$ moves on $O A$, and $Q$ moves on $O B$ while maintaining $|\overrightarrow{P Q}|$ as a constant $a\left(P, Q\right.$ do not coincide with $O$). Given $\angle A O B=60^{\circ}$, $a=\sqrt{7}$, then the range of $\frac{\overrightarrow{P Q} \cdot \overrightarrow{P O}}{|\overrightarrow{P O}|}+\frac{3 \overrightarrow{Q P} \cdot \overrightarrow{Q O}}{|\overrightarrow{Q O}|}$ is $\qquad$.
(10) Let $\angle O P Q=\alpha$, then $\angle O Q P=120^{\circ}-\alpha$, therefore
$$
\begin{aligned}
& \frac{\overrightarrow{P Q} \cdot \overrightarrow{P O}}{|\overrightarrow{P O}|}+\frac{3 \overrightarrow{Q P} \cdot \overrightarrow{Q O}}{|\overrightarrow{Q O}|} \\
= & a \cos \alpha+3 a \cos \left(120^{\circ}-\alpha\right) \\
= & a\left(\cos \alpha+3 \cos 120^{\circ} \cdot \cos \alpha+3 \sin 120^{\circ} \cdot \sin \alpha\right) \\
= & a\left(-\frac{1}{2} \cos \alpha+\frac{3 \sqrt{3}}{2} \sin \alpha\right) \\
= & -\frac{1}{2} a(\cos \alpha-3 \sqrt{3} \sin \alpha) \\
= & -\sqrt{7} a \cos (\alpha+\arctan 3 \sqrt{3}) \\
= & -7 \cos (\alpha+\arctan 3 \sqrt{3}),
\end{aligned}
$$

When $0^{\circ}<\alpha<120^{\circ}$,
$$
\cos (\alpha+\arctan 3 \sqrt{3}) \in\left[-1, \frac{\sqrt{7}}{14}\right),
$$

Therefore, the range of the original expression is $\left(-\frac{\sqrt{7}}{2}, 7\right]$.