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