+433 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 $k$.
+130071 Let P = (x,y)
PA^2 + PB^2 + PC^2 = 3PQ^2 + k
(x -2)^2 + (y -4)^2 + ( x + 3)^2 + (y -1)^2 + (x -1)^2 + ( y - 7)^2 = 3PQ^2 + k
3x^2 + 3y^2 + 24y + 80 = 3PQ^2 + k complete the square on y
3x^2 + 3(y^2 + 8y + 80/3) = 3PQ^2 + k
3x^2 + 3( y^2 + 8y + 16 + 80/3 - 16) = 3PQ^2 + k
3x^2 + 3(y + 4)^2 + 3 ( 80/3 - 48/3) = 3PQ^2 + k
3x^2 + 3(y + 4)^2 + 3 (32/3) = 3PQ^2 + k
3 [ x^2 + (y + 4)^2 ] + 32 = 3 PQ^2 + k
P =(x, y) Q = (0, -4)
PQ^2 = [ x^2 + ( y + 4)^2]
k = 32
