Let ${\mathrm{z}}_1,{\mathrm{z}}_2,{\mathrm{z}}_3\in \mathrm{C}$ such that ${\mathrm{z}}_1+{\mathrm{z}}_2+{\mathrm{z}}_3=0\text{ and }\left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|=\left|{\mathrm{z}}_3\right|=1\text{ then }\left|{\mathrm{z}}_1^2+2{\mathrm{z}}_2^2+{\mathrm{z}}_3^2\right|=$

$$\begin{gathered} {\mathrm{z}}_1+{\mathrm{z}}_2+{\mathrm{z}}_3=0\Rightarrow \frac{{\mathrm{z}}_1+{\mathrm{z}}_2+{\mathrm{z}}_3}{3}=0 \\ \therefore \mathrm{Centroid} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{formed} \mathrm{by} \mathrm{the} \mathrm{vertices} {\mathrm{z}}_1,{\mathrm{z}}_2,{\mathrm{z}}_3 \mathrm{is} \mathrm{at} \mathrm{origin}. \\ \left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|=\left|{\mathrm{z}}_3\right|=1\Rightarrow \mathrm{circumcentre} \mathrm{is} \mathrm{at} \mathrm{origin} \\ \mathrm{centroid}= \mathrm{circumcentre} \end{gathered}$$.The triangle is equilateral with R = 1 with side = $\sqrt{3}$.
$\begin{array}{l}{{\Rightarrow {\mathrm{z}}_1^2+2{\mathrm{z}}_2^2+\mathrm{z}}_3}^2={\left({\mathrm{z}}_1+{\mathrm{z}}_3\right)}^2+2{\mathrm{z}}_2^2-2{\mathrm{z}}_1{\mathrm{z}}_3\\ =3{\mathrm{z}}_2^2-2{\mathrm{z}}_2^2={\mathrm{z}}_2^2\\ \left|{\mathrm{z}}_1^2+2{\mathrm{z}}_2^2+{\mathrm{z}}_3^2\right|=\left|{\mathrm{z}}_2^2\right|={\left|z_2\right|}^2=1\end{array}$