Hard Math geometry problem

By the distance formula,

$$\begin{align*} PA^2 &= (x - 4)^2 + (y + 4)^2, \ PB^2 &= (x - 3)^2 + (y - 5)^2, \ PC^2 &= (x + 1)^2 + (y - 2)^2. \end{align*}$$

Hence, the given equation becomes $$\begin{align*} (x - 4)^2 + (y + 4)^2 + (x - 3)^2 + (y - 5)^2 + (x + 1)^2 + (y - 2)^2 &= 3((x - Q)^2 + (y - Q)^2) + k \ \Rightarrow \quad 3x^2 + 3y^2 - 14x - 14y + 12 &= 3x^2 - 6Qx + 3Q^2 + 3y^2 - 6Qy + 3Q^2 + k \ \Rightarrow \quad -14x - 14y &= -6Qx - 6Qy + k - 12 \ \Rightarrow \quad 7x + 7y &= 3Q(x + y) + \frac{k - 12}{7}. \end{align*}$$

Since this equation holds for all points P, the coefficients of x and y must be zero.

Hence, Q=−7/3 and k−12=0, so k = 12​.