'I' is the incentre of triangle ABC whose corresponding sides are a, b, c, respectively then IA→+bIB→+cIC→  is always equal to

Let the incentre be at the origin and be A(p→),B(q→) and C(r→) then IA→=p→,IB→=q→ and IC→=r→Incentre I is  ap→+bq→+cr→a+b+c  where p = BC, q=AC and r=AB Incentre is at the origin. Therefore, ap→+bq→+cr→a+b+c=0→, or ap→+bq→+cr→=0→⇒aIA→+bIB→+cIC→=0→