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

• A
  $3$
• B
  $2$
• C
  $\frac{2}{3}$
• D
  $\frac{3}{4}$