$\bar{i}+\bar{j}+\bar{k}, a_1 \bar{i}+b_1 \bar{j}+c_1 \bar{k}, a_2 \bar{i}+b_2 \bar{j}+c_2 \bar{k}, a_3 \bar{i}+b_3 \bar{j}+c_3 \bar{k}$ are the position vectors of the points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ respectively. $\frac{2}{3}(\bar{i}+\bar{j}+\bar{k})$ is the position vector of the centroid of the triangular face BCD of the tetrahedron ABCD. If $\alpha \bar{i}+\beta \bar{j}+\gamma \bar{k}$ is the position vector of the centroid of the tetrahedron, then $2 \alpha+\beta+\gamma=$

• Correct answer
  A

  3
• B

  2
• C

  $\frac{2}{3}$
• D

  $\frac{3}{4}$

Hence, the correct answer is option: 'A'.