The volume of a tetrahedron whose vertices are $A \equiv (-1, 2, 3)$,$B \equiv (3, -2, 1)$,$C \equiv (2, 1, 3)$,and $D \equiv (-1, -2, 4)$ is

If the volume of the parallelepiped formed by the vectors $\hat{i} + \lambda \hat{j} + \hat{k}$,$\hat{j} + \lambda \hat{k}$ and $\lambda \hat{i} + \hat{k}$ is minimum,then $\lambda$ is equal to

The volume of the parallelepiped determined by the vectors $\vec{a} + \vec{b}, \vec{b} + \vec{c}$ and $\vec{c} + \vec{a}$ is $4$. Then the volume of the parallelepiped determined by the vectors $\vec{a} \times \vec{b}, \vec{b} \times \vec{c}$ and $\vec{c} \times \vec{a}$ is:

If $a=2u+3v+7w$,$b=u+v-2w$ and $c=-u-2v-3w$,then $\left|\frac{[u, v, w]}{[a, b, c]}\right|(a+b+c) = $

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b}$,and $\vec{c} = \hat{j} - \hat{k}$ be three vectors such that $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 1$. If the length of the projection vector of the vector $\vec{b}$ on the vector $\vec{a} \times \vec{c}$ is $l$,then the value of $3l^{2}$ is equal to $.....$

Let $\vec V = 2\hat i + \hat j - \hat k$,$\vec W = \hat i + 3\hat k$,and $|\vec U| = 2$. If $\vec U$ is a vector in the $x-y$ plane,then the greatest value of $([\vec U \vec V \vec W])^2$ is: