Let the three sides of a triangle ABC be given by the vectors $$2\hat{i} - \hat{j} + \hat{k}$$, $$\hat{i} - 3\hat{j} - 5\hat{k}$$ and $$3\hat{i} - 4\hat{j} - 4\hat{k}$$. Let G be the centroid of the triangle ABC. Then $$6\left(|\vec{AG}|^2 + |\vec{BG}|^2 + |\vec{CG}|^2\right)$$ is equal to ______.

Given the side vectors of triangle $$ABC$$ be:

$$\vec{AB} = 2\hat{i} - \hat{j} + \hat{k}$$

$$\vec{BC} = \hat{i} - 3\hat{j} - 5\hat{k}$$

$$\vec{AC} = \vec{AB} + \vec{BC} = 3\hat{i} - 4\hat{j} - 4\hat{k}$$

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Calculating the squares of the lengths of the sides of the triangle:

$$c^2 = |\vec{AB}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6$$

$$a^2 = |\vec{BC}|^2 = 1^2 + (-3)^2 + (-5)^2 = 1 + 9 + 25 = 35$$

$$b^2 = |\vec{AC}|^2 = 3^2 + (-4)^2 + (-4)^2 = 9 + 16 + 16 = 41$$

The sum of the squares of the sides is:

$$a^2 + b^2 + c^2 = 35 + 41 + 6 = 82$$

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Using the centroid property for any triangle $$ABC$$ with centroid $$G$$:

$$|\vec{AG}|^2 + |\vec{BG}|^2 + |\vec{CG}|^2 = \frac{1}{3}\left(a^2 + b^2 + c^2\right)$$

Substituting the value of the sum of the squares of the sides:

$$|\vec{AG}|^2 + |\vec{BG}|^2 + |\vec{CG}|^2 = \frac{82}{3}$$

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Multiplying the expression by 6 to find the required value:

$$6\left(|\vec{AG}|^2 + |\vec{BG}|^2 + |\vec{CG}|^2\right) = 6 \times \frac{82}{3} = 2 \times 82 = 164$$

Therefore, the final value is equal to 164.