ok actually i did it with coordinate bash ..............pls if anyone has another shorter approach send it

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We use the following theorem: Let $P$ be any point in the plane of a ∆ $ABC$. Then $$(PA)^2 + (PB)^2 + (PC)^2 = (GA)^2 + (GB)^2 + (GC)^2 + 3(PG)^2$$. This can be proved for example by assigning complex coordinates and using distance formula. Here $G$ denotes the centroid of the triangle. For this problem, $P$ is a point on the concerned circle and $G = O$ where $O$ is the center of the circle. Also as a consequence of Apollonius's Theorem, $(GA)^2 + (GB)^2 + (GC)^2=1/3(a^2+b^2+c^2)$, where $a,b,c$ denote the sides of the triangle $ABC$ In this case, since $ABC$ is equilateral, we have $(PA)^2 + (PB)^2 + (PC)^2 = a^2$ where $a$ is the side length. And $3(PG)^2 = 3R^2 = 3a^2$ (which can be shown by elementary geometry). Thus the required sum of squares of distances is $4a^2$, which is a constant. Q.E.D