If origin is the orthocenter of a triangle formed by the points  $\left(\text{cosα,sinα,0}\right)\text{,}\left(\text{cosβ,sinβ,0}\right),\left(\text{cosγ,sinγ,0}\right) Then \sum \text{cos}\left(\text{2α-β-γ}\right)=\_\_\_$

$$\begin{gathered} \text{OA= OB = OC }\Rightarrow \text{circumcentres\hspace{0.17em}= O(0,0)} \\ And also orthocenter =0(0,0,0)=G \\ \Delta \text{ABC\hspace{0.17em}is equilateral }\therefore \cos \alpha +\cos \beta +\cos \gamma =0\&\sin \alpha +\sin \beta +\sin \gamma =0 \\ \Rightarrow \cos \alpha +\cos \beta =-\cos \gamma \&\sin \alpha +\sin \beta =-\sin \gamma Sq. and adding \\ \Rightarrow \cos (\alpha -\beta )=\frac{-1}{2}\text{\hspace{0.17em} IIrly cos }\left(\text{β - γ}\right)\text{= cos}\left(\text{γ- α}\right)\text{=}\frac{\text{-1}}{\text{2}} \\ \cos (2\alpha -\beta -\gamma )=\cos [(\alpha -\beta )-(\gamma -\alpha )]=1 \Rightarrow \sum \cos (2\alpha -\beta -\gamma )=1+1+1=3 \end{gathered}$$