If circumcentre of an equilateral triangle inscribed in $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ with vertices having eccentric angles $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }$ respectively is $\left(x_1,y_1\right)$ then $\mathrm{\Sigma Cos}\mathrm{\alpha }\cdot \cos \mathrm{\beta }+\sum \sin \mathrm{\alpha sin}\mathrm{\beta }$ is

A(a cos$\mathrm{\alpha }$ , b sin$\mathrm{\alpha }$), B($\mathrm{\alpha }$ Cos $\mathrm{\beta }$, b sin $\mathrm{\beta }$), C($\mathrm{\alpha }$ cos r, b sin r) centroid = circumcentre
$\begin{array}{l}\left({\mathrm{x}}_{1,}{\mathrm{y}}_1\right)=\left(\frac{\sum \mathrm{\alpha cos}\mathrm{\alpha }}{3},\frac{\sum \mathrm{bsin}\mathrm{\alpha }}{3}\right)\\ \Rightarrow \frac{3{\mathrm{x}}_1}{\mathrm{a}}=\sum \cos \mathrm{\alpha },\frac{3{\mathrm{y}}_1}{\mathrm{b}}=\sum \sin \mathrm{\alpha }\\ \Rightarrow \left(\frac{9{\mathrm{x}}_1^2}{{\mathrm{a}}^2}+\frac{9_1^2}{{\mathrm{b}}^2}-3\right)=-2\left(\sum \cos \mathrm{\alpha sin}\mathrm{\beta }\frac{9{\mathrm{x}}^2}{2{\mathrm{a}}^2}+\frac{9{\mathrm{y}}^2}{2{\mathrm{b}}^2}-\frac{3}{2}\right.\end{array}$