If $A(-1, 2, 3)$,$B(3, -2, 1)$,$C(2, 1, 3)$ and $D(-1, -2, 4)$ are the vertices of a tetrahedron,then its volume is

If $a=2 \hat{i}-3 \hat{j}+5 \hat{k}$, $b=3 \hat{i}-4 \hat{j}+5 \hat{k}$ and $c=5 \hat{i}-3 \hat{j}-2 \hat{k}$, then the volume of the parallelepiped with coterminous edges $a+b$, $b+c$, $c+a$ is

If $A \equiv (1, -1, 0)$,$B \equiv (0, 1, -1)$,and $C \equiv (-1, 0, 1)$,then the unit vector $\overline{d}$ such that $\overline{a}$ and $\overline{d}$ are perpendicular and $\overline{b}, \overline{c}, \overline{d}$ are coplanar is

Three lines are drawn from the origin $O$ with direction ratios proportional to $(1, -1, 1)$, $(2, -3, 0)$, and $(1, 0, 3)$. The three lines are

The altitude of the parallelepiped, whose coterminous edges are the vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 4\hat{j} - \hat{k}$, and $\vec{c} = \hat{i} + \hat{j} + 3\hat{k}$, where $\vec{a}$ and $\vec{b}$ are the sides of the base of the parallelepiped, is

$a$ is perpendicular to both $b$ and $c$. The angle between $b$ and $c$ is $\frac{2 \pi}{3}$. If $|a|=2$, $|b|=3$, and $|c|=4$, then $c \cdot (a \times b)$ is equal to (in $\sqrt{3}$)