If $\overline{a}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k}$,$\overline{b}=4 \hat{\imath}+5 \hat{\jmath}+3 \hat{k}$ and $\overline{c}=6 \hat{\imath}+\hat{\jmath}+5 \hat{k}$ are the position vectors of the vertices of a triangle $ABC$ respectively,then the position vector of the intersection of the medians (centroid) of the triangle $ABC$ is:

  • A
    $4 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}$
  • B
    $2 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}$
  • C
    $5 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}$
  • D
    $3 \hat{\imath}+3 \hat{\jmath}+4 \hat{k}$

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