Vector Algebra

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