$\displaystyle\frac{z_1}{z_2}+\frac{z_2}{z_3}+\frac{z_3}{z_1} \in\mathbb{R}\implies \displaystyle\frac{z_1}{z_2}+\frac{z_2}{z_3}+\frac{z_3}{z_1}=\overline{\displaystyle\frac{z_1}{z_2}+\frac{z_2}{z_3}+\frac{z_3}{z_1}}\implies \displaystyle\frac{\overline{z_1}}{\overline{z_2}}+\frac{\overline{z_2}}{\overline{z_3}}+\frac{\overline{z_3}}{\overline{z_1}} $ $|z_1|=|z_2|=|z_3| \implies \overline{z_i} = \frac{1}{z_i} \implies \displaystyle\frac{z_1}{z_2}+\frac{z_2}{z_3}+\frac{z_3}{z_1} = \displaystyle\frac{z_2}{z_1}+\frac{z_3}{z_2}+\frac{z_1}{z_3}$ It follows $(z_1-z_2)(z_2-z_3)(z_3-z_1)=0$ If $z_1=z_2$, then $\displaystyle\frac{z_3}{z_1}=\pm i \implies |az_1 + bz_2 + cz_3| = |z_1|\cdot|a+b\pm ci|=\sqrt{(a+b)^2 + c^2}$ If $z_2 = z_3$, then $|az_1 + bz_2 + cz_3| = \sqrt{(b+c)^2 + a^2}$. If $z_3 = z_1$, then $|az_1 + bz_2 + cz_3| = \sqrt{(c+a)^2 + b^2}$.