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

### Core Logic Let the vertices of the triangle be A$A$, B$B$, and C$C$. The vectors representing the side paths are: overlineAB = 2hati - hatj + hatk$$\overline{AB} = 2\hat{i} - \hat{j} + \hat{k}$$ overlineCA = hati - 3hatj - 5hatk$$\overline{CA} = \hat{i} - 3\hat{j} - 5\hat{k}$$ overlineCB = 3hati - 4hatj - 4hatk$$\overline{CB} = 3\hat{i} - 4\hat{j} - 4\hat{k}$$ Notice that overlineAB + overlineCA = (2+1)hati + (-1-3)hatj + (1-5)hatk = 3hati - 4hatj - 4hatk = overlineCB$\overline{AB} + \overline{CA} = (2+1)\hat{i} + (-1-3)\hat{j} + (1-5)\hat{k} = 3\hat{i} - 4\hat{j} - 4\hat{k} = \overline{CB}$. This structurally validates vector addition rules.

### Step 1: Finding Position Vectors relative to A Let's set vertex A$A$ as the origin origin point (Position vector vecA = vec0$\vec{A} = \vec{0}$): - Position vector of B$B$: vecB = 2hati - hatj + hatk$\vec{B} = 2\hat{i} - \hat{j} + \hat{k}$ - Position vector of C$C$: Since overlineCA = vecA - vecC = -vecC implies vecC = -hati + 3hatj + 5hatk$\overline{CA} = \vec{A} - \vec{C} = -\vec{C} \implies \vec{C} = -\hat{i} + 3\hat{j} + 5hat{k}$ Now, calculate the position vector of the centroid G$G$: vecG = fracvecA + vecB + vecC3 = fracvec0 + (2hati - hatj + hatk) + (-hati + 3hatj + 5hatk)3 = frac13left(hati + 2hatj + 6hatkright)$$\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} = \frac{\vec{0} + (2\hat{i} - \hat{j} + \hat{k}) + (-\hat{i} + 3\hat{j} + 5\hat{k})}{3} = \frac{1}{3}\left(\hat{i} + 2\hat{j} + 6\hat{k}\right)$$ ### Step 2: Calculating Squared Lengths to the Centroid Let's find each individual vector distance block: - overlineAG = vecG - vecA = frac13(hati + 2hatj + 6hatk) implies |overlineAG|^2 = frac19(1^2 + 2^2 + 6^2) = frac419$\overline{AG} = \vec{G} - \vec{A} = \frac{1}{3}(\hat{i} + 2\hat{j} + 6\hat{k}) \implies |\overline{AG}|^2 = \frac{1}{9}(1^2 + 2^2 + 6^2) = \frac{41}{9}$ - overlineBG = vecG - vecB = left(frac13-2right)hati + left(frac23+1right)hatj + left(2-1right)hatk = -frac53hati + frac53hatj + 1hatk$\overline{BG} = \vec{G} - \vec{B} = \left(\frac{1}{3}-2\right)\hat{i} + \left(\frac{2}{3}+1\right)\hat{j} + \left(2-1\right)\hat{k} = -\frac{5}{3}\hat{i} + \frac{5}{3}\hat{j} + 1\hat{k}$ |overlineBG|^2 = left(-frac53right)^2 + left(frac53right)^2 + 1^2 = frac259 + frac259 + 1 = frac599$$|\overline{BG}|^2 = \left(-\frac{5}{3}\right)^2 + \left(\frac{5}{3}\right)^2 + 1^2 = \frac{25}{9} + \frac{25}{9} + 1 = \frac{59}{9}$$ - overlineCG = vecG - vecC = left(frac13+1right)hati + left(frac23-3right)hatj + left(2-5right)hatk = frac43hati - frac73hatj - 3hatk$\overline{CG} = \vec{G} - \vec{C} = \left(\frac{1}{3}+1\right)\hat{i} + \left(\frac{2}{3}-3\right)\hat{j} + \left(2-5\right)\hat{k} = \frac{4}{3}\hat{i} - \frac{7}{3}\hat{j} - 3\hat{k}$ |overlineCG|^2 = left(frac43right)^2 + left(-frac73right)^2 + (-3)^2 = frac169 + frac499 + 9 = frac1469$$|\overline{CG}|^2 = \left(\frac{4}{3}\right)^2 + \left(-\frac{7}{3}\right)^2 + (-3)^2 = \frac{16}{9} + \frac{49}{9} + 9 = \frac{146}{9}$$ ### Step 3: Final Targeted Evaluation Summing the squared values and multiplying by 6$6$: textValue = 6 left[ |overlineAG|^2 + |overlineBG|^2 + |overlineCG|^2 right] = 6 left[ frac419 + frac599 + frac1469 right]$$\text{Value} = 6 \left[ |\overline{AG}|^2 + |\overline{BG}|^2 + |\overline{CG}|^2 \right] = 6 \left[ \frac{41}{9} + \frac{59}{9} + \frac{146}{9} \right]$$ textValue = 6 times frac2469 = 2 times frac2463 = 2 times 82 = 164$$\text{Value} = 6 \times \frac{246}{9} = 2 \times \frac{246}{3} = 2 \times 82 = 164$$ ### Pattern Recognition Setting one vector node as the origin point (vecA = vec0$\vec{A} = \vec{0}$) heavily dampens intermediate coordinate math steps, avoiding dealing with an absolute baseline origin orientation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra