If $\vec p$ and $\vec q$ are unit vectors such that $[\vec p, \vec q, \vec p \times \vec q] = \frac{1}{2}$,then the angle between $\vec p$ and $\vec q$ is

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

Let a vector $\vec{a}$ be coplanar with vectors $\vec{b}=2 \hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+\hat{k}$. If $\vec{a}$ is perpendicular to $\vec{d}=3 \hat{i}+2 \hat{j}+6 \hat{k}$,and $|\vec{a}|=\sqrt{10}$. Then a possible value of $[\vec{a} \vec{b} \vec{c}]+[\vec{a} \vec{b} \vec{d}]+[\vec{a} \vec{c} \vec{d}]$ is equal to:

Let the volume of the tetrahedron with vertices $\hat{i}-\hat{j}-2\hat{k}$, $-2\hat{i}+\hat{j}-2\hat{k}$, $-\hat{i}-2\hat{j}+\hat{k}$, and $2\hat{i}+2\hat{j}+a\hat{k}$ be $\frac{20}{3}$. Then the integral value of $a$ is

The number of distinct real values of $\lambda$,for which the vectors $-\lambda^2 \hat{i}+\hat{j}+\hat{k}$,$\hat{i}-\lambda^2 \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}-\lambda^2 \hat{k}$ are coplanar,is

For any non-zero vectors $a, b, c$,$a \cdot[(b+c) \times(a+b+c)] = \ldots .$