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If $\hat{i}$ is the position vector of the centroid $G$ of triangle $\mathrm{ABC}$ and $2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ are respectively the position vectors of its vertices $\mathrm{A}$ and $\mathrm{B}$, then $\mathrm{AG}^2+\mathrm{BG}^2+\mathrm{CG}^2=$
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Verified Answer
The correct answer is:
$74$
Let vertex of $c$ is $(x \hat{i}+y \hat{j}+z \hat{k})$
$\begin{aligned}
& \therefore \text { Centroid } G=i+0 \hat{j}+0 . \hat{k} \\
& =\frac{(2 \hat{i}+\hat{j}+\hat{k})+(2 \hat{i}+4 \hat{j}-4 \hat{k})+(x \hat{i}+y \hat{j}+z \hat{k})}{3}
\end{aligned}$
$\begin{aligned} & \Rightarrow 3=4+x \Rightarrow x=-1 \\ & \text { or } 0=5+y \Rightarrow y=-5 \\ & \text { or } 0=-3+z \Rightarrow z=3, \text { so, } c(-1,-5,3) \\ & \text { Now, } \mathrm{AG}^2+\mathrm{BG}^2+\mathrm{CG}^2 \\ & =\left(1^2+1^2+1^2\right)+\left(1^2+4^2+4^2\right)+\left(2^2+5^2+3^2\right) \\ & =3+33+38=74\end{aligned}$
$\begin{aligned}
& \therefore \text { Centroid } G=i+0 \hat{j}+0 . \hat{k} \\
& =\frac{(2 \hat{i}+\hat{j}+\hat{k})+(2 \hat{i}+4 \hat{j}-4 \hat{k})+(x \hat{i}+y \hat{j}+z \hat{k})}{3}
\end{aligned}$
$\begin{aligned} & \Rightarrow 3=4+x \Rightarrow x=-1 \\ & \text { or } 0=5+y \Rightarrow y=-5 \\ & \text { or } 0=-3+z \Rightarrow z=3, \text { so, } c(-1,-5,3) \\ & \text { Now, } \mathrm{AG}^2+\mathrm{BG}^2+\mathrm{CG}^2 \\ & =\left(1^2+1^2+1^2\right)+\left(1^2+4^2+4^2\right)+\left(2^2+5^2+3^2\right) \\ & =3+33+38=74\end{aligned}$
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