Let the vectors $\bar{A B} = 2 \bar{i} + 2 \bar{j} + \bar{k}$ and $\bar{A C} = 2 \bar{i} + 4 \bar{j} + 4 \bar{k}$ be two sides of a triangle ABC. If G is the centroid of $Δ A B C$ ,then $\frac{22}{7} {(\bar{A G})}^{2} + 5 =$

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answer is B.

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Detailed Solution

Take A as reference point
$\bar{A G} = \frac{\bar{O O} + \bar{O B} + \bar{O C}}{3} = \frac{4 i + 6 j + 5 k}{3}$
$|\bar{A G}| = \frac{\sqrt{77}}{3} \Rightarrow \frac{27}{7} {|\bar{A G}|}^{2} + 5 = 38$

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