If $z_0$ is the circumcenter of an equilateral triangle with vertices $z_1 ,z_2 , z_3$, then  $\frac{{z_1}^2+{z_2}^2 +{z_3}^2}{{z_0}^2}=----$

Since z₁, z₂, z₃, are vertices of an equilateral triangle, so 
${\mathrm{z}}_1^2+{\mathrm{z}}_2^2+{\mathrm{z}}_3^2={\mathrm{z}}_1{\mathrm{z}}_2+{\mathrm{z}}_2{\mathrm{z}}_3+{\mathrm{z}}_3{\mathrm{z}}_1$
Further the circumcenter of an equilateral triangle is same as its centroid, so
$\begin{array}{l}{\mathrm{z}}_0=\left({\mathrm{z}}_1+{\mathrm{z}}_2+{\mathrm{z}}_3\right)/3\Rightarrow 9{\mathrm{z}}_0^2={\mathrm{z}}_1^2+{\mathrm{z}}_2^2+{\mathrm{z}}_3^2\\ +2\left({\mathrm{z}}_1{\mathrm{z}}_2+{\mathrm{z}}_2{\mathrm{z}}_3+{\mathrm{z}}_3{\mathrm{z}}_1\right)={\mathrm{z}}_1^2+{\mathrm{z}}_2^2+{\mathrm{z}}_3^2+2\left({\mathrm{z}}_1^2+{\mathrm{z}}_2^2+{\mathrm{z}}_3^2\right)\\ \Rightarrow {\mathrm{z}}_1^2+{\mathrm{z}}_2^2+{\mathrm{z}}_3^2=3{\mathrm{z}}_0^2.\end{array}$