Školjka

%V0 Let $Q$ be the centre of the inscribed circle of a triangle $ABC.$ Prove that for any point $P,$ $$a(PA)^2 + b(PB)^2 + c(PC)^2 = a(QA)^2 + b(QB)^2 + c(QC)^2 + (a + b + c)(QP)^2,$$ where $a = BC, b = CA$ and $c = AB.$