+9479 We are given:
A = (4, 1)
B = (6, 2)
C = (-1, 2)
And we can let:
P = (x, y)
By the Pythagorean Theorem/distance formula we can say:
PA2 = (x - 4)2 + (y - 1)2
PB2 = (x - 6)2 + (y - 2)2
PC2 = (x + 1)2 + (y - 2)2
Then....
PA2 + PB2 + PC2 = (x - 4)2 + (y - 1)2 + (x - 6)2 + (y - 2)2 + (x + 1)2 + (y - 2)2
Expand each term, then combine like terms to get:
PA2 + PB2 + PC2 = 3x2 - 18x + 3y2 - 10y + 62
(I used this to do that because I am lazy.)
Next we want to get the expression on the right side of the equation into the form:
3PQ2 + k
which is:
3[ (x - something)2 + (y - something)2 ] + k
To do that, we need to complete the squares of the x terms and the y terms.
PA2 + PB2 + PC2 = 3(x2 - 6x) + 3(y2 - \(\frac{10}{3}\)y) + 62
PA2 + PB2 + PC2 = 3(x2 - 6x + 9 - 9) + 3(y2 - \(\frac{10}{3}\)y + \(\frac{25}{9}\) - \(\frac{25}{9}\)) + 62
PA2 + PB2 + PC2 = 3( (x - 3)2 - 9) + 3( (y - \(\frac53\))2 - \(\frac{25}{9}\)) + 62
PA2 + PB2 + PC2 = 3(x - 3)2 - 27 + 3(y - \(\frac53\))2 - \(\frac{25}{3}\) + 62
PA2 + PB2 + PC2 = 3(x - 3)2 + 3(y - \(\frac53\))2 - \(\frac{25}{3}\) + 62 - 27
PA2 + PB2 + PC2 = 3[ (x - 3)2 + (y - \(\frac53\))2 ] - \(\frac{25}{3}\) + 62 - 27
PA2 + PB2 + PC2 = 3[ (x - 3)2 + (y - \(\frac53\))2 ] + \(\frac{80}{3}\)
(Check)
Now it is in the desired form, and so we can pick out that k = \(\frac{80}{3}\)
BTW, I came across this answer. It works out a bit more cleanly if A = (4, -1). Just wanted to mention this in case that is what you meant. 
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