$A\left(z_1\right),B\left(z_2\right),C\left(z_3\right)$  are the vertices of an equilateral triangle ABC, whose circumcentre is $D\left(z_0\right),$  then $z_1^2+z_2^2+z_3^2$  is always equal to

$A\left(z_1\right),B\left(z_2\right),C\left(z_3\right)$   form an equilateral triangle so $z_1^2+z_2^2+z_3^2=z_1{z}_2+z_2{z}_3+z_3{z}_1$

Also, $\Delta ABC$ is equilateral, so circumcentre $z_0$ is also centroid.

            $z_1+z_2+z_3=3z_0$

$\Rightarrow$      $z_1^2+z_2^2+z_3^2+2\left(z_1{z}_2+z_2{z}_3+z_3{z}_1\right)=9z_0^2$

$\Rightarrow$     $z_1^2+z_2^2+z_3^2=3z_0^2$