$\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}}, a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}+c_2 \hat{\mathbf{k}}, a_3 \hat{\mathbf{i}}+b_3 \hat{\mathbf{j}}+c_3 \hat{\mathbf{k}}$ 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 $2 \alpha+\beta+\gamma=$

If $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{b}=9 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-18 \hat{\mathbf{k}}$ are two vectors, then $\frac{\text { Projection of } \mathbf{b} \text { on } \mathbf{a}}{\text { Projection of } \mathbf{a} \text { on } \mathbf{b}}=$

Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ be three vectors. If $\mathbf{r}$ is a vector such that $\mathbf{r} \cdot \mathbf{a}=0$, $\mathbf{r} \cdot \mathbf{b}=-2$ and $\mathbf{r} \cdot \mathbf{c}=6$, then $\mathbf{r} \cdot(\beta \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=$

Let $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{c}=6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ be three vectors. If $\mathbf{d}$ is a vector perpendicular to both $\mathbf{a}, \mathbf{b}$ and $|\mathbf{d} \times \mathbf{c}|=14$, then $|\mathbf{d} \cdot \mathbf{c}|=$