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 $$6\bigl(|\overrightarrow{AG}|^2+|\overrightarrow{BG}|^2+|\overrightarrow{CG}|^2\bigr)$$ is equal to:

Step 1: Verify if vectors form a triangle

First, we need to check if the given vectors can form a triangle. For three vectors to form a triangle, the sum of any two vectors must be equal to the third vector (or their sum must be the zero vector if they are taken in order around the triangle). Let the given vectors be $\vec{v}_{1} = 2\hat{i}-\hat{j}+\hat{k}$, $\vec{v}_{2} = \hat{i}-3\hat{j}-5\hat{k}$, and $\vec{v}_{3} = 3\hat{i}-4\hat{j}-4\hat{k}$. We observe that $\vec{v}_{1} + \vec{v}_{2} = (2\hat{i}-\hat{j}+\hat{k}) + (\hat{i}-3\hat{j}-5\hat{k}) = (2+1)\hat{i} + (-1-3)\hat{j} + (1-5)\hat{k} = 3\hat{i}-4\hat{j}-4\hat{k}$. This sum is equal to $\vec{v}_{3}$. Therefore, these three vectors can represent the sides of a triangle.

Step 2: Recall the centroid property

For a triangle ABC with centroid G, there is a known property relating the sum of the squares of the distances from the vertices to the centroid and the sum of the squares of the side lengths. This property states that the sum of the squares of the lengths of the medians is equal to three-fourths the sum of the squares of the sides. More directly, the sum of the squares of the distances from the vertices to the centroid is one-third the sum of the squares of the side lengths.

Step 3: Calculate the square of the magnitudes of the side vectors

Now, we calculate the square of the magnitude for each given vector. The magnitude of a vector $a\hat{i}+b\hat{j}+c\hat{k}$ is $\sqrt{a^2+b^2+c^2}$, so its square is $a^2+b^2+c^2$. We apply this formula to each of the three vectors.

Step 4: Substitute values into the centroid property

We substitute the calculated squared magnitudes of the side vectors into the centroid property formula. The sum of the squares of the side lengths is $6 + 35 + 41 = 82$. Therefore, the sum of the squares of the distances from the vertices to the centroid is $\frac{1}{3} \times 82$.

Step 5: Calculate the final expression

Finally, we need to find the value of $6\bigl(|\overrightarrow{AG}|^2+|\overrightarrow{BG}|^2+|\overrightarrow{CG}|^2\bigr)$. We multiply the result from the previous step by 6. This simplifies to $2 \times 82 = 164$.

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