Since the equation PA^2 + PB^2+PC^2=3PQ^2 + k holds true for any point P, we may let P = A, P = B, P = C and get 3 equations.
Let Q = (r, s). Then substituting P = A, B, C respectively gives: {53+65=3((r−4)2+(s+1)2)+k53+146=3((r+3)2+(s−1)2)+k65+146=3((r−8)2+(s−6)2)+k.
Solving this system is possible, but it is very tedious work, so I will do it with Wolfram Alpha:
By the Wolfram Alpha computation, k = 88, r = 3, s = 2.
In case you're wondering how to solve the system of equations, let (1) be the first equation, (2) be the second equation, (3) be the third equation. Consider (1) - (2), (2) - (3), and (1) - (3).