Answer:
k = 46
Step-by-step explanation:
Let P = (x,y). Then
PA^2 + PB^2 + PC^2 &= (x - 4)^2 + (y + 1)^2 + (x + 3)^2 + (y - 1)^2 + (x - 5)^2 + (y + 3)^2
= (x^2 - 8x + 16) + (y^2 + 2y + 1) + (x^2 + 6x + 9) + (y^2 - 2y + 1) + (x^2 - 10x + 25) + (y^2 + 6y + 9)
= 3x^2 - 12x + 3y^2 + 6y + 61.
Completing the square in $x$ and $y$, we get
PA^2 + PB^2 + PC^2 &= 3(x^2 - 4x) + 3(y^2 + 2y) + 61
= 3(x^2 - 4x + 4) - 3*4 + 3(y^2 + 2y + 1) - 3*1 + 61
= 3(x - 2)^2 + 3(y + 1)^2 + 46.
Thus, Q = (2,-1), and
PA^2 + PB^2 + PC^2 = 3PQ^2 + 46, so the constant k is 46.
We see that Q is the centroid of triangle ABC because the coordinates of Q are the average of the coordinates of the vertices of ABC. From this result, we can conclude that PA^2 + PB^2 + PC^2 is minimized when P is the centroid and that the minimum value is 46.