If $z_1,z_2,z_3$ are three distinct complex numbers and  $a,b,c$ are there positive real numbers such that $\frac{a}{|z_2-z_3|}=\frac{b}{\left|z_3-z_1\right|)}=\frac{c}{\left|z_1-z_2\right|}$ then $\frac{a^2}{\left(z_2-z_3\right)}+\frac{b^2}{(z_3-z_1)}+\frac{c^2}{(z_1-z_2)}=$ 

Let $\frac{a}{\left|z_2-z_3\right|}=\frac{b}{\left|z_3-z_1\right|}=\frac{c}{\left|z_1-z_2\right|}=\lambda$  (say)

$\Rightarrow a=\lambda \left|z_2-z_3\right|,b=\lambda \left|z_3-z_1\right|,c=\lambda \left|z_1-z_2\right|$

$\Rightarrow a^2={\lambda }^2{\left|z_2-z_3\right|}^2={\lambda }^2\left(z_2-z_3\right)\left({\overline{z}}_2-{\overline{z}}_3\right)$

$\therefore \frac{a^2}{\left(z_2-z_3\right)}={\lambda }^2\left({\overline{z}}_2-{\overline{z}}_3\right)$

Similarly, $\frac{b^2}{\left(z_3-z_1\right)}=\lambda \left({\overline{z}}_3-{\overline{z}}_1\right)$ and $\frac{c^2}{\left|z_1-z_2\right|}={\lambda }^2\left({\overline{z}}_1-{\overline{z}}_2\right)$

$\therefore \frac{a^2}{z_2-z_3}+\frac{b^2}{z_3-z_1}+\frac{c^2}{z_1-z_2}=0$ .