If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} - 4\hat{k}$,and $\vec{c} = \hat{i} + \lambda \hat{j} + 3\hat{k}$ are coplanar,then the value of $\lambda$ is:

The volume of the tetrahedron having the edges $\hat{i}+2\hat{j}-\hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, and $\hat{i}-\hat{j}+\lambda\hat{k}$ as coterminous edges is $\frac{2}{3}$ cubic units. Then $\lambda$ equals:

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

If the points with position vectors $\hat{i}-\hat{j}+\hat{k}$, $2 \hat{i}-\hat{k}$, $\hat{j}+2 \hat{k}$ and $\hat{i}+\hat{j}+\lambda \hat{k}$ are coplanar, then the magnitude of the vector $6 \lambda \hat{i}-3 \hat{j}+6 \hat{k}$ is

If $\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vectors and $\vec{r}$ is any vector,then $[\vec{b} \, \vec{c} \, \vec{r}] \vec{a} + [\vec{c} \, \vec{a} \, \vec{r}] \vec{b} + [\vec{a} \, \vec{b} \, \vec{r}] \vec{c} = \dots$

$[(\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c})] \cdot \vec{d} = \dots$