+448 Points $A,$ $B,$ and $C$ are given in the coordinate plane. There exists a point $Q$ and a constant $k$ such that for any point $P$,
PA^2 + PB^2 + PC^2 = 3PQ^2 + k.
If $A = (2,4),$ $B = (-3,1),$ and $C = (1,7)$, then find the constant $k$.
+130458 Let P = (x,y)
PA^2 + PB^2 + PC^2 =
(x -2)^2 + ( y- 4)^2 + (x + 3)^2 + (y -1)^2 + (x-1)^2 + ( y-7)^2 ....simplifying, we have
3x^2 +3y^2 -24y + 80 complete the square on y
3[ (x-0)^2 + y^2 - 24y + 144] + 80 - 3(144) =
3 [ ( x-0)^2 + (y- 12)^2 ] - 352 = 3PQ^2 + k
Q= (0 , 12)
k = -352
