[Solution] Let $e^{i \theta}=\cos \theta+i \sin \theta$. Suppose
$$\frac{z_{1}}{z_{2}}=e^{i \theta}, \frac{z_{2}}{z_{3}}=e^{i \varphi},$$

then $\quad \frac{z_{3}}{z_{1}}=e^{-i(\theta+\varphi)}$.
From the given condition, we have
$$e^{i \theta}+e^{i \varphi}+e^{-i(\theta+\varphi)}=1$$

Taking the imaginary part on both sides, we get
$$\begin{array}{l} \sin \theta+\sin \varphi-\sin (\theta+\varphi)=0 \\ 2 \sin \frac{\theta+\varphi}{2} \cos \frac{\theta-\varphi}{2}-2 \sin \frac{\theta+\varphi}{2} \cos \frac{\theta+\varphi}{2}=0 \\ \sin \frac{\theta+\varphi}{2} \cdot \sin \frac{\theta}{2} \sin \frac{\varphi}{2}=0 \end{array}$$

Therefore, we have
$\theta=2 k \pi$ or $\varphi=2 k \pi$ or $\theta+\varphi=2 k \pi, k \in Z$. Hence, $z_{1}=z_{2}$ or $z_{2}=$ $z_{3}$ or $z_{3}=z_{1}$.

If $z_{1}=z_{2}$, substituting into the given equation, we get
$$\begin{array}{l} 1+\frac{z_{1}}{z_{3}}+\frac{z_{3}}{z_{1}}=1 \\ \left(\frac{z_{1}}{z_{3}}\right)^{2}=-1 \\ \frac{z_{1}}{z_{3}}= \pm i \end{array}$$

Thus, we have
$$\left|a z_{1}+b z_{2}+c z_{3}\right|=\left|z_{1}\right| \cdot|a+b \pm c i|=\sqrt{(a+b)^{2}+c^{2}} .$$

Similarly, if $z_{2}=z_{3}$, then
$$\left|a z_{1}+b z_{2}+c z_{3}\right|=\sqrt{a^{2}+(b+c)^{2}}$$

If $z_{3}=z_{1}$, then
$$\left|a z_{1}+b z_{2}+c z_{3}\right|=\sqrt{(a+c)^{2}+b^{2}}$$

The required value is $\sqrt{(a+b)^{2}+c^{2}}$ or $\sqrt{a^{2}+(b+c)^{2}}$ or $\sqrt{(c+a)^{2}+b^{2}}$.