Key Concepts and Formulas

• Centroid Property: For a triangle $ABC$ with centroid $G$, the sum of the squares of the distances from the centroid to the vertices is related to the squares of the side lengths by the formula: $|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2 = \frac{1}{3}(a^2+b^2+c^2)$ where $a = |\overrightarrow{BC}|$, $b = |\overrightarrow{CA}|$, and $c = |\overrightarrow{AB}|$ are the lengths of the sides opposite to vertices $A, B, C$ respectively.
• Vector Magnitude Squared: The square of the magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $|\vec{v}|^2 = x^2 + y^2 + z^2$.
• Triangle Law of Vector Addition: If $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are two vectors representing two sides of a triangle, then the third side is represented by $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$.

Step-by-Step Solution

Step 1: Understand the Goal and Simplify the Target Expression The problem asks for the value of $6\left(|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right)$. Using the key concept relating the centroid to the side lengths, we can rewrite this expression. From the centroid property, we have: $|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2 = \frac{1}{3}(a^2+b^2+c^2)$ Therefore, the expression we need to calculate becomes: $6\left(|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right) = 6 \times \frac{1}{3}(a^2+b^2+c^2) = 2(a^2+b^2+c^2)$ Our primary task is to find the sum of the squares of the side lengths of the triangle, $a^2+b^2+c^2$, from the given vectors.

Step 2: Interpret the Given Vectors as Sides of the Triangle Let the three given vectors be $\vec{u} = 2 \hat{i}-\hat{j}+\hat{k}$, $\vec{v} = \hat{i}-3 \hat{j}-5 \hat{k}$, and $\vec{w} = 3 \hat{i}-4 \hat{j}-4 \hat{k}$. For these vectors to form the sides of a triangle, they must satisfy a vector addition relationship. Typically, two vectors originating from a common vertex sum up to the third side. Let's check if any pair of vectors sums to the third. Consider the sum of $\vec{u}$ and $\vec{v}$: $\vec{u} + \vec{v} = (2 \hat{i}-\hat{j}+\hat{k}) + (\hat{i}-3 \hat{j}-5 \hat{k})$ $\vec{u} + \vec{v} = (2+1)\hat{i} + (-1-3)\hat{j} + (1-5)\hat{k}$ $\vec{u} + \vec{v} = 3\hat{i} - 4\hat{j} - 4\hat{k}$ We observe that $\vec{u} + \vec{v} = \vec{w}$. This relationship implies that if we consider $\vec{u}$ as $\overrightarrow{AB}$ and $\vec{v}$ as $\overrightarrow{BC}$, then by the triangle law of vector addition, $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \vec{u} + \vec{v}$. Since $\vec{u} + \vec{v} = \vec{w}$, we have $\overrightarrow{AC} = \vec{w}$. Thus, the three sides of the triangle $ABC$ can be represented by the vectors:

• $\overrightarrow{AB} = \vec{u} = 2 \hat{i}-\hat{j}+\hat{k}$
• $\overrightarrow{BC} = \vec{v} = \hat{i}-3 \hat{j}-5 \hat{k}$
• $\overrightarrow{AC} = \vec{w} = 3 \hat{i}-4 \hat{j}-4 \hat{k}$

Step 3: Calculate the Squares of the Side Lengths Now we calculate the square of the magnitude of each of these side vectors. These will correspond to $c^2$, $a^2$, and $b^2$ respectively, based on our assignment in Step 2.

• The square of the length of side $AB$ (opposite vertex $C$) is $c^2$: $c^2 = |\overrightarrow{AB}|^2 = |\vec{u}|^2 = (2)^2 + (-1)^2 + (1)^2 = 4 + 1 + 1 = 6$
• The square of the length of side $BC$ (opposite vertex $A$) is $a^2$: $a^2 = |\overrightarrow{BC}|^2 = |\vec{v}|^2 = (1)^2 + (-3)^2 + (-5)^2 = 1 + 9 + 25 = 35$
• The square of the length of side $AC$ (opposite vertex $B$) is $b^2$: $b^2 = |\overrightarrow{AC}|^2 = |\vec{w}|^2 = (3)^2 + (-4)^2 + (-4)^2 = 9 + 16 + 16 = 41$

Step 4: Calculate the Sum of the Squares of the Side Lengths We sum the squares of the side lengths calculated in Step 3: $a^2+b^2+c^2 = 35 + 41 + 6 = 82$

Step 5: Compute the Final Expression Substitute the sum of the squares of the side lengths into the simplified expression derived in Step 1: $6\left(|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right) = 2(a^2+b^2+c^2)$ $= 2(82)$ $= 164$

Common Mistakes & Tips

• Incorrect Vector Interpretation: Assuming the given vectors are $\overrightarrow{AB}, \overrightarrow{BC}, \overrightarrow{CA}$ directly without checking their sum. The vectors may represent sides originating from a common vertex or other configurations. Always verify the vector addition property.
• Calculation Errors: Mistakes in squaring components or summing them up can lead to incorrect side lengths. Double-check magnitude calculations.
• Misapplication of Centroid Formula: Ensure the formula relates the sum of squared distances from the centroid to the vertices with the sum of squared side lengths, not vertex coordinates directly.

Summary The problem required calculating $6\left(|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right)$. We used the property that for a triangle $ABC$ with centroid $G$, $3(|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2) = a^2+b^2+c^2$. This simplified the target expression to $2(a^2+b^2+c^2)$. We identified that the given vectors $\vec{u} = 2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{v} = \hat{i}-3 \hat{j}-5 \hat{k}$ summed to $\vec{w} = 3 \hat{i}-4 \hat{j}-4 \hat{k}$, which allowed us to define the sides of the triangle as $\overrightarrow{AB}=\vec{u}$, $\overrightarrow{BC}=\vec{v}$, and $\overrightarrow{AC}=\vec{w}$. We then calculated the squares of the lengths of these sides: $|\overrightarrow{AB}|^2=6$, $|\overrightarrow{BC}|^2=35$, and $|\overrightarrow{AC}|^2=41$. The sum of these squares is $6+35+41=82$. Finally, we computed $2(82) = 164$.

The final answer is $\boxed{164}$.