If $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$,and $\vec{c} = \lambda\hat{i} + \hat{j} + (2\lambda - 1)\hat{k}$ are coplanar vectors,then $\lambda = . . . .$

If $\overline{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \overline{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$ and $\overline{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$ are non-zero non-coplanar vectors and $m$ is a non-zero scalar such that $[m\overline{a}+\overline{b} \quad m\overline{b}+\overline{c} \quad m\overline{c}+\overline{a}] = 28[\overline{a} \quad \overline{b} \quad \overline{c}]$,then the value of $m$ is:

$a \cdot (a \times b) = $

Let $\vec{a}=\hat{i}-\alpha \hat{j}+\beta \hat{k}$,$\vec{b}=3 \hat{i}+\beta \hat{j}-\alpha \hat{k}$ and $\vec{c}=-\alpha \hat{i}-2 \hat{j}+\hat{k}$,where $\alpha$ and $\beta$ are integers. If $\vec{a} \cdot \vec{b}=-1$ and $\vec{b} \cdot \vec{c}=10$,then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to $.....$

If the position vectors of the vertices $A, B$ and $C$ are $6i$,$6j$ and $k$ respectively with respect to the origin $O$,then the volume of the tetrahedron $OABC$ is

If $a = i + j + k$,$b = 4i + 3j + 4k$,and $c = i + \alpha j + \beta k$ are coplanar vectors and $|c| = \sqrt{3}$,then: