Let $z_1,z_2,z_3$ be complex numbers of equal magnitude satisfying $z_1+z_2+z_3=-\frac{\sqrt{3}}{2}-i\sqrt{5}$ and $z_1{z}_2{z}_3=\sqrt{3}+i\sqrt{5}.$  If $z_1=x_1+i{y}_1,z_2=x_2+i{y}_2,z_3=x_3+i{y}_3;x_1,x_2,x_3,y_1,y_2,y_3\in R$ then $\frac{16{(x_1{y}_1+x_2{y}_2+x_3{y}_3)}^2}{5}=$

$$\begin{gathered} \left|z_1{z}_2{z}_3\right|=2\sqrt{2}so\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=\sqrt{2} \\ Now, {\overline{z}}_1+{\overline{z}}_2+{\overline{z}}_3=-\frac{\sqrt{3}}{2}+i\sqrt{5} \\ \Rightarrow \frac{2(z_1{z}_2+z_2{z}_3+z_3{z}_1)}{z_1{z}_2{z}_3}=-\frac{\sqrt{3}}{2}+i\sqrt{5} \\ \Rightarrow z_1{z}_2+z_2{z}_3+z_3{z}_1=(-\frac{\sqrt{3}}{4}+\frac{i\sqrt{5}}{2})(\sqrt{3}+i\sqrt{5}) \\ Now, x_1{y}_1+x_2{y}_2+x_3{y}_3=\frac{\mathrm{Im}(z_1^2+z_2^2+z_3^2)}{2} \\ =\frac{I_m{(z_1+z_2+z_3)}^2}{2}-\mathrm{Im}(z_1{z}_2+z_2{z}_3+z_3{z}_1) \\ =\frac{\sqrt{15}}{2}-(-\frac{\sqrt{15}}{4}+\frac{\sqrt{15}}{2})=\frac{\sqrt{15}}{4} \end{gathered}$$