Let $a, b$ and $c$ be three vectors. Then the scalar triple product $[a, b, c]$ is equal to:

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + \hat{k}$,and $\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$,then the value of $\left| \begin{matrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c} \end{matrix} \right|$ is

The volume of the tetrahedron with $\hat{i}-\lambda \hat{j}+\hat{k}$, $\lambda \hat{i}-\hat{j}-\hat{k}$ and $\hat{i}+\hat{j}+\lambda \hat{k}$ as coterminous edges is $2$. If $\lambda$ is an integer, then $|\lambda \hat{i}-3 \lambda \hat{j}+3 \hat{k}|=$

The parallelepiped is determined by vectors $\vec{a} = -2\hat{i} + 5\hat{j} + 3\hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}$, and $\vec{c} = -3\hat{i} + \hat{j} + 4\hat{k}$. The altitude of the parallelepiped on the parallelogram base determined by vectors $\vec{b}$ and $\vec{c}$ is

If $a$ is perpendicular to $b$ and $c$,$|a| = 2$,$|b| = 3$,$|c| = 4$ and the angle between $b$ and $c$ is $\frac{2\pi}{3}$,then $[a \; b \; c]$ is equal to (in $\sqrt{3}$):

If $[\vec{p}-\vec{r}, \vec{q}, \vec{s}] + [\vec{p}+\vec{q}, \vec{r}, \vec{s}] = m[\vec{p}, \vec{r}, \vec{s}] + n[\vec{q}, \vec{r}, \vec{s}] + t[\vec{p}, \vec{q}, \vec{s}]$,then the values of $m$,$n$,$t$ respectively are . . . . . .