Two ships leave a port at the same time. One of them move in the direction of $E 50^{\circ} \mathrm{N}$ with a speed of 8 kmph and the other moves in the direction of $\mathrm{S} 20^{\circ} \mathrm{E}$ with a speed of 12 kmph . Then, the distance between the ships at the end of 2 h is (in km )

In a $\triangle A B C$, if $\mathbf{B C}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{C A}=6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$, then the perimeter of the triangle is

$\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}}, a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}+c_2 \hat{\mathbf{k}}, a_3 \hat{\mathbf{i}}+b_3 \hat{\mathbf{j}}+c_3 \hat{\mathbf{k}}$ 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 $\frac{\text { Projection of } \mathbf{b} \text { on } \mathbf{a}}{\text { Projection of } \mathbf{a} \text { on } \mathbf{b}}=$