If origin is the orthocentre of the triangle with vertices $A(\cos \alpha ,\sin \alpha )$, $B(\cos \beta ,\sin \beta )$ and $C(\cos \gamma ,\sin \gamma )$ then $\cos (2\alpha -\beta -\gamma )+$ $\cos (2\beta -\gamma -\alpha )+$$\cos (2\gamma -\alpha -\beta )=$

OA = OB = OC

Circumcentre  S = (0,0)

Orthocentre O = (0,0)

Triangle is equilateral

$\left(\frac{{\displaystyle \cos \alpha +\cos \beta +\cos \gamma }}{{\displaystyle 3}},\frac{{\displaystyle \sin \alpha +\sin \beta +\sin \gamma }}{{\displaystyle 3}}\right)=(0,0)$

$\cos \alpha +\cos \beta +\cos \gamma =0----\left(1\right)$

$\sin \alpha +\sin \beta +\sin \gamma =0----\left(2\right)$

$$\begin{gathered} x=\cos \alpha +i \sin \alpha , y=\cos \beta +i\sin \beta , z=\cos \gamma +i\sin \gamma \\ x+y+z=0 \\ {x}^3+y^3+z^3=3xyz \\ \frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}=3 \\ \sum _{ }\cos (2\alpha -\beta -\gamma )=3 \end{gathered}$$