$$\frac{3}{|z_2-z_3|} = \frac{4}{|z_3-z_1|}=\frac{5}{|z_1-z_2|}$$ Find the value of $$\frac{9}{|z_2-z_3|}+\frac{16}{|z_3-z_1|}+\frac{25}{|z_1-z_2|}$$
Let The given equality be equal to some value k Then $$9=k^2|z_2-z_3|^2$$ and so on
Replacing these values $$k^2[|z_2-z_3|+|z_3-z_1|+|z_1-z_2|]$$ How should I solve further? The minimum value can be calculated to be zero, but the question is asking for the actual value.