If $\hat{i}$ is the position vector of the centroid $G$ of triangle $ABC$ and $2\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}-4\hat{k}$ are respectively the position vectors of its vertices $A$ and $B$,then $AG^2+BG^2+CG^2=$

  • A
    $77$
  • B
    $74$
  • C
    $86$
  • D
    $83$

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$I$. Two non-zero, non-collinear vectors are linearly independent.
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