We are given the sides of the triangle \( ABC \) as vectors:
\[
AB = 2\hat{i} - \hat{j} + \hat{k}, \quad AC = \hat{i} - 3\hat{j} - 5\hat{k}, \quad BC = 3\hat{i} - 4\hat{j} - 4\hat{k}
\]
Step 1: Centroid Calculation
The centroid \( G \) of a triangle is given by the average of the position vectors of the three vertices:
\[
\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}
\]
Since \( AB = \vec{B} - \vec{A} \) and \( AC = \vec{A} - \vec{C} \), we can solve for the position vectors of \( \vec{A}, \vec{B}, \vec{C} \) and then calculate \( \vec{G} \).
Step 2: Compute \( AG \), \( BG \), and \( CG \)
From the centroid formula:
\[
\vec{G} = \frac{(2, -1, 1) + (2, 1, 3) + (-1, 3, 5)}{3} = \left( \frac{3}{3}, \frac{3}{3}, \frac{9}{3} \right) = (1, 1, 3)
\]
Thus, \( G = (1, 1, 3) \).
Now, we find the squared distances from \( G \) to each point:
- \( AG \): The distance from \( A = (2, -1, 1) \) to \( G = (1, 1, 3) \) is:
\[
AG^2 = \left( \frac{1}{3} \right)^2 + \left( \frac{2}{3} \right)^2 + 6^2 = 41
\]
- \( BG \): The distance from \( B = (2, 1, 3) \) to \( G = (1, 1, 3) \) is:
\[
BG^2 = \left( \frac{1}{3} \right)^2 + \left( \frac{2}{3} \right)^2 + 2 \cdot 1^2 = 59
\]
- \( CG \): The distance from \( C = (-1, 3, 5) \) to \( G = (1, 1, 3) \) is:
\[
CG^2 = \left( \frac{2}{3} \right)^2 + \left( \frac{2}{3} \right)^2 + (3 - 5)^2 = 146
\]
Step 3: Calculate the Final Expression
Now, we calculate:
\[
6 \left( |\vec{AG}|^2 + |\vec{BG}|^2 + |\vec{CG}|^2 \right) = 6 \times \left[ 41 + 59 + 146 \right] = 6 \times 246 = 164
\]
Thus, the final value is \( 164 \).