Question 1202365
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Let P be (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 =

= 3x^2 + 3y^2 - 12x + 6y + 61 = 

=3(x^2 + y^2 - 4x + 2y) + 61 =

= 3((x-2)^2 + (y+1)^2) + 46.


This shows that if  Q = (2,-1),  then  PQ^2 = (x-2)^2 + (y+1)^2.

And so  PA^2 + PB^2 + PC^2 = 3PQ^2 + 46.


<U>ANSWER</U>.  k = 46.
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Solved.