Example 7 Given real numbers $a, b, c$. It is known that 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 the value of $\left|a z_{1}+b z_{2}+c z_{3}\right|$.
From the given, we know that $z_{1}=\frac{1}{\bar{z}_{1}}$, etc., so $\frac{\bar{z}_{2}}{\bar{z}_{1}}+\frac{\bar{z}_{3}}{\bar{z}_{2}}+\frac{\bar{z}_{1}}{\bar{z}_{3}}=1$, which means $\frac{z_{2}}{z_{1}}+\frac{z_{3}}{z_{2}}+\frac{z_{1}}{z_{3}}=1$. This can be rewritten as
$$
\frac{z_{1}}{z_{2}} \cdot \frac{z_{2}}{z_{3}}+\frac{z_{1}}{z_{2}} \cdot \frac{z_{3}}{z_{1}}+\frac{z_{2}}{z_{3}} \cdot \frac{z_{3}}{z_{1}}=1 \text {, and } \frac{z_{1}}{z_{2}} \cdot \frac{z_{2}}{z_{3}} \cdot \frac{z_{3}}{z_{1}}=1 \text {. }
$$

Thus, $\frac{z_{1}}{z_{2}}, \frac{z_{2}}{z_{3}}, \frac{z_{3}}{z_{1}}$ are the three roots of the equation $x^{3}-x^{2}+x-1=0$. Therefore, $\left\{\frac{z_{1}}{z_{2}}, \frac{z_{2}}{z_{3}}, \frac{z_{3}}{z_{1}}\right\}=\{1, \mathrm{i},-\mathrm{i}\}$. Without loss of generality, let $\frac{z_{1}}{z_{2}}=1, \frac{z_{2}}{z_{3}}=\mathrm{i}, \frac{z_{3}}{z_{1}}=-\mathrm{i}$. Then $z_{2}=z_{1}, z_{3}=-z_{1} \mathrm{i}$.
Thus,
$$
\begin{array}{c}
\left|a z_{1}+b z_{2}+c z_{3}\right|=\left|a z_{1}+b z_{1}-c z_{1} \mathrm{i}\right| \\
=|a+b-c \mathrm{i}|=\sqrt{(a+b)^{2}+c^{2}} .
\end{array}
$$

By symmetry, the desired value is $\sqrt{(a+b)^{2}+c^{2}}$, or $\sqrt{(a+c)^{2}+b^{2}}, \sqrt{(b+c)^{2}+a^{2}}$.