How does one solve this problem?

Given points $A = (4, -4)$, $B = (3, 8)$, and $C = (-1, 2)$, and a point $Q$ such that for any point $P$ in the plane, the following equation holds:

 

$$PA^2 + PB^2 + PC^2 = 3PQ^2 + k$$

 

We need to find the constant $k$.

 

### Step 1: Write down the expression for the sum of squares of distances

Let $P = (x, y)$ and $Q = (x_Q, y_Q)$. The squared distances from $P$ to the points $A$, $B$, and $C$ are:

 

$$PA^2 = (x - 4)^2 + (y + 4)^2$$

$$PB^2 = (x - 3)^2 + (y - 8)^2$$

$$PC^2 = (x + 1)^2 + (y - 2)^2$$

 

The sum of these squared distances is:

 

$$PA^2 + PB^2 + PC^2 = \left[(x - 4)^2 + (y + 4)^2\right] + \left[(x - 3)^2 + (y - 8)^2\right] + \left[(x + 1)^2 + (y - 2)^2\right]$$

 

Expanding these expressions:

 

$$PA^2 = (x^2 - 8x + 16) + (y^2 + 8y + 16)$$

$$PB^2 = (x^2 - 6x + 9) + (y^2 - 16y + 64)$$

$$PC^2 = (x^2 + 2x + 1) + (y^2 - 4y + 4)$$

 

Adding them together:

 

$$PA^2 + PB^2 + PC^2 = \left[3x^2 + (-8x - 6x + 2x) + (16 + 9 + 1)\right] + \left[3y^2 + (8y - 16y - 4y) + (16 + 64 + 4)\right]$$

 

Simplifying:

 

$$PA^2 + PB^2 + PC^2 = 3x^2 - 12x + 26 + 3y^2 - 12y + 84$$

 

Thus:

 

$$PA^2 + PB^2 + PC^2 = 3(x^2 - 4x + \frac{26}{3}) + 3(y^2 - 4y + 28)$$

 

### Step 2: Express the right-hand side

The expression for $3PQ^2$ is:

 

$$3PQ^2 = 3\left[(x - x_Q)^2 + (y - y_Q)^2\right] = 3\left[(x^2 - 2xx_Q + x_Q^2) + (y^2 - 2yy_Q + y_Q^2)\right]$$

 

Expanding:

 

$$3PQ^2 = 3x^2 - 6xx_Q + 3x_Q^2 + 3y^2 - 6yy_Q + 3y_Q^2$$

 

### Step 3: Set up the equation

We equate $PA^2 + PB^2 + PC^2$ with $3PQ^2 + k$:

 

$$3x^2 - 12x + 110 + 3y^2 - 12y + 84 = 3x^2 - 6xx_Q + 3x_Q^2 + 3y^2 - 6yy_Q + 3y_Q^2 + k$$

 

By comparing coefficients, we get:

 

$$-12x = -6xx_Q \quad \text{and} \quad -12y = -6yy_Q$$

 

So:

 

$$x_Q = 2, \quad y_Q = 2$$

 

Now, matching the constant terms:

 

$$110 + 84 = 3x_Q^2 + 3y_Q^2 + k$$

$$194 = 3(2^2) + 3(2^2) + k = 12 + 12 + k = 24 + k$$

 

Thus:

$$k = 194 - 24 = 170$$

 

### Final Answer:

The constant $k$ is $\boxed{170}$.