ETV $* \star$ Given real numbers $a, b, c$. It is known that the complex numbers $z_{1}, z_{2}, z_{3}$ satisfy
$$
\left\{\begin{array}{l}
\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1, \\
\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{3}}+\frac{z_{3}}{z_{1}}=1 .
\end{array}\right.
$$

Find $\left|a z_{1}+b z_{2}+c z_{3}\right|$.
Let $p=\frac{z_{1}}{z_{2}}, q=\frac{z_{2}}{z_{3}}$, then from (1) we know $|p|=|q|=1$. Substituting into (2), we get $p+q+\frac{1}{p q}=1$, i.e., $p+q+\overline{p q}=1$. Thus, we have $\sin \alpha+\sin \beta-\sin (\alpha+\beta)=0$, where $\alpha=\arg p, \beta=\arg q$. From (3), we get $\sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\alpha+\beta}{2}=0$. Therefore, $\frac{\alpha}{2}, \frac{\beta}{2}, \frac{\alpha+\beta}{2}$ must have one that is 0 or $\pi$, i.e., $p, q$ or $p q$ must have one that is a real number. Without loss of generality, assume $p \in \mathbf{R}$. Then, from $|p|=1$, we know $p=1$ or -1.

If $p=-1$, then $q-q=2$, which is impossible. Therefore, $p=1$, at this time $q+\bar{q}=0$. Combining $|q|=1$, we get $q= \pm \mathrm{i}$. Thus,
$$
\begin{aligned}
\left|a z_{1}+b z_{2}+c z_{3}\right| & =\left|a \frac{z_{1}}{z_{2}}+b+c \frac{z_{3}}{z_{2}}\right|=|a p+b+c \bar{q}| \\
& =|a+b \pm c \mathrm{i}|=\sqrt{(a+b)^{2}+c^{2}} .
\end{aligned}
$$

By symmetry, there are two other possibilities. Therefore, $\left|a z_{1}+b z_{2}+c z_{3}\right|=\sqrt{(a+b)^{2}+c^{2}}$ or $\sqrt{a^{2}+(b+c)^{2}}$ or $\sqrt{b^{2}+(a+c)^{2}}$.