Let $\vec{OD} = \hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{CB} = -3\hat{i} - 2\hat{k}$ be the diagonals of the parallelogram $OBDC$. If $\vec{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$, then the volume of the parallelepiped determined by vectors $\vec{OA}, \vec{OB}$, and $\vec{OC}$ (in cubic units) is:

The value of $a$,so that the volume of the parallelepiped formed by $\hat{i} + a \hat{j} + \hat{k}$,$\hat{j} + a \hat{k}$,and $a \hat{i} + \hat{k}$ becomes minimum is

If the volume of a parallelepiped,whose coterminous edges are given by the vectors $\overrightarrow{a} = \hat{i} + \hat{j} + n\hat{k}$,$\overrightarrow{b} = 2\hat{i} + 4\hat{j} - n\hat{k}$,and $\overrightarrow{c} = \hat{i} + n\hat{j} + 3\hat{k}$ $(n \geq 0)$,is $158$ cubic units,then which of the following is true?

Consider $\overrightarrow{r}, \overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ are non-zero vectors such that $\overrightarrow{r} \cdot \overrightarrow{a}=0$,$|\overrightarrow{r} \times \overrightarrow{b}|=|\overrightarrow{r}||\overrightarrow{b}|$,and $|\overrightarrow{r} \times \overrightarrow{c}|=|\overrightarrow{r}||\overrightarrow{c}|$. Then,the scalar triple product $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$ is:

$(a+b) \cdot(b+c) \times(a+b+c)$ is equal to

Volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is $12$ cubic units. The volume of a tetrahedron with coterminous edges $\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{a} + \vec{b} - \vec{c}$ will be ............. cubic units.