10. (20 points) Given complex numbers $z_{1}, z_{2}, z_{3}$ with arguments $\alpha, \beta, \gamma$ respectively, and satisfying:
$$
\left|z_{1}\right|=1,\left|z_{2}\right|+\left|z_{3}\right|=2, z_{1}+z_{2}+z_{3}=0 .
$$

Find the range of $\cos (\alpha-\beta)+2 \cos (\beta-\gamma)+3 \cos (\gamma-\alpha)$.
10. Since $z_{2}=-\left(z_{1}+z_{3}\right)$, we have
$$
\begin{array}{l}
\left|z_{2}\right|=\left|z_{1}+z_{3}\right|=2-\left|z_{3}\right| . \\
\text { By }\left|z_{3}\right|-\left|z_{1}\right| \leqslant\left|z_{1}+z_{3}\right| \leqslant\left|z_{1}\right|+\left|z_{3}\right| \\
\Rightarrow\left|z_{3}\right|-1 \leqslant 2-\left|z_{3}\right| \leqslant 1+\left|z_{3}\right| \\
\Rightarrow \frac{1}{2} \leqslant\left|z_{3}\right| \leqslant \frac{3}{2} .
\end{array}
$$

Similarly, $\frac{1}{2} \leqslant\left|z_{2}\right| \leqslant \frac{3}{2}$.
Let $\left|z_{2}\right|=k \in\left[\frac{1}{2}, \frac{3}{2}\right],\left|z_{3}\right|=2-k$. Then $\cos \alpha+k \cos \beta+(2-k) \cos \gamma=0$, $\sin \alpha+k \sin \beta+(2-k) \sin \gamma=0$.
Combining $\cos ^{2} \alpha+\sin ^{2} \alpha=1$, eliminating $\alpha$ we get
$$
\cos (\beta-\gamma)=\frac{2 k^{2}-4 k+3}{2 k(k-2)} \text {. }
$$

Similarly, $\cos (\gamma-\alpha)=\frac{5-4 k}{2(k-2)}$,
$$
\cos (\alpha-\beta)=\frac{3-4 k}{2 k} \text {. }
$$

Therefore, the original expression $=-6+\frac{-3}{k-2} \in[-4,0]$.