lf P is any arbitrary point on the circumcircle of the equilateral triangle of side length L  units, then |PA→|2+|PB→|2+|PC→|2  is always equal to

Let P.V. of P, A,B and C be  p→,a→,b→ and c→  respectively, and O(0→) be the circumcentre of equilateral triangle ABC. Then |p→|=|b→|=|a→|=|c→|=l3 Now |PA→|2=|a→−p→|2=|a→|2+|p→|2−2p→⋅a→ Similarly, |PB→|2=|b→|2+|p→|2−2p→⋅b→ and  |PC→|2=|c→|2+|p→|2−2p→⋅c→⇒ Σ|PA→|2=6⋅l23−2p→⋅(a→+b→+c→)=2l2 ( as a→+b→+c→/3=0→)