Let \( A = (4, -4) \), \( B = (3, 8) \), and \( C = (-1, 2) \). We need to find \( k \) such that the equation
\[ PA^2 + PB^2 + PC^2 = 3PQ^2 + k \]
holds for a point \( P \) and a point \( Q \), for any point \( P \).
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Calculate \( PA^2 + PB^2 + PC^2 \): The squares of distances from point \( P = (x, y) \) to points \( A \), \( B \), and \( C \) are given by:
- \( PA^2 = (x - 4)^2 + (y + 4)^2 \)
- \( PB^2 = (x - 3)^2 + (y - 8)^2 \)
- \( PC^2 = (x + 1)^2 + (y - 2)^2 \)
Expanding each square:
\[ PA^2 = (x - 4)^2 + (y + 4)^2 = (x^2 - 8x + 16) + (y^2 + 8y + 16) = x^2 + y^2 - 8x + 8y + 32 \]
\[ PB^2 = (x - 3)^2 + (y - 8)^2 = (x^2 - 6x + 9) + (y^2 - 16y + 64) = x^2 + y^2 - 6x - 16y + 73 \]
\[ PC^2 = (x + 1)^2 + (y - 2)^2 = (x^2 + 2x + 1) + (y^2 - 4y + 4) = x^2 + y^2 + 2x - 4y + 5 \]
Adding these three distances:
\[ PA^2 + PB^2 + PC^2 = (x^2 + y^2 - 8x + 8y + 32) + (x^2 + y^2 - 6x - 16y + 73) + (x^2 + y^2 + 2x - 4y + 5) \]
Combining like terms gives:
\[ PA^2 + PB^2 + PC^2 = 3x^2 + 3y^2 + (-8 - 6 + 2)x + (8 - 16 - 4)y + (32 + 73 + 5) \] Simplifying the coefficients:
\[ = 3x^2 + 3y^2 - 12x - 12y + 110 \]
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Next, we compute \( 3PQ^2 + k \): Let \( Q = (q_x, q_y) \). Then \( PQ^2 = (x - q_x)^2 + (y - q_y)^2 \) which expands to:
\[ PQ^2 = (x^2 - 2qq_x + q_x^2) + (y^2 - 2qq_y + q_y^2) = x^2 + y^2 - 2qx + q_x^2 + q_y^2 \]
Therefore,
\[ 3PQ^2 = 3(x^2 + y^2 - 2qq_x - 2qq_y + q_x^2 + q_y^2) = 3x^2 + 3y^2 - 6qx - 6qy + 3(q_x^2 + q_y^2) \]
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Setting the equations equal:
\[ 3x^2 + 3y^2 - 12x - 12y + 110 = 3x^2 + 3y^2 - 6qx - 6qy + 3(q_x^2 + q_y^2) + k \]
By matching coefficients for \( x \) and \( y \):
From the coefficients of \( x \):
\[ -12 = -6q \implies q = 2 \]
From the coefficients of \( y \):
\[ -12 = -6r \implies r = 2 \]
Thus, \( Q = (2, 2) \).
Now substituting \( Q \) back to find \( k \):
With \( q = 2 \):
\[ \begin{align*} 3(q_x^2 + q_y^2) &= 3(2^2 + 2^2) = 3(4 + 4) = 3 \cdot 8 = 24 \ k &= 110 - 24 = 86 \end{align*} \]
Thus, the value of \( k \) is
\[ \boxed{86}. \]