2. 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 the value of $\left|a z_{1}+b z_{2}+c z_{3}\right|$.
Let $\alpha=\frac{z_{1}}{z_{2}}, \beta=\frac{z_{2}}{z_{3}}, \gamma=\frac{z_{3}}{z_{1}}$, then $\alpha+\beta+\gamma=1, \alpha \beta+\beta \gamma+\gamma \alpha=1, \alpha \beta \gamma=1$. By Vieta's formulas, $\alpha, \beta, \gamma$ are the three roots of the cubic equation $x^{3}-x^{2}+x-1=\left(x^{2}+1\right)(x-1)=0$, and they are distinct. If $\alpha=\frac{z_{1}}{z_{2}}=1, \frac{z_{2}}{z_{3}}= \pm \mathrm{i}$, then $\left|a z_{1}+b z_{2}+c z_{3}\right|=\frac{1}{\left|z_{2}\right|} \cdot\left|a+b+c \cdot \frac{z_{3}}{z_{2}}\right|=|a+b \mp c \mathrm{i}|=\sqrt{(a+b)^{2}+c^{2}} ;$ if $\beta=\frac{z_{2}}{z_{3}}=1, \frac{z_{3}}{z_{1}}= \pm \mathrm{i}$, then $\left|a z_{1}+b z_{2}+c z_{3}\right|=\frac{1}{\left|z_{3}\right|} \cdot\left|b+c+a \cdot \frac{z_{1}}{z_{3}}\right|=|b+c \mp a \mathrm{i}|=\sqrt{(b+c)^{2}+a^{2}} ;$ if $\gamma=\frac{z_{3}}{z_{1}}=1, \frac{z_{1}}{z_{2}}= \pm \mathrm{i}$, then $\left|a z_{1}+b z_{2}+c z_{3}\right|=\frac{1}{\left|z_{1}\right|} \cdot\left|a+c+b \cdot \frac{z_{2}}{z_{1}}\right|=|a+c \mp b \mathrm{i}|=\sqrt{(a+c)^{2}+b^{2}}$. In summary, $\left|a z_{1}+b z_{2}+c z_{3}\right|=\sqrt{(a+b)^{2}+c^{2}}$ or $\sqrt{(b+c)^{2}+a^{2}}$ or $\sqrt{(a+c)^{2}+b^{2}}$.