The volume of the tetrahedron whose vertices are A$(-1, 2, 3)$, B$(3, -2, 1)$, C$(p, 1, 3)$, D$(-1, -2, 4)$ is $\dfrac{16}{3}$ cubic units then the value of p is

Let $\overline{OD} = \hat{i} + 2\hat{j} + 6\hat{k}$, $\overline{CB} = -3\hat{i} - 2\hat{k}$ be the diagonals of the parallelogram OBDC and $\overline{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$ be another vector. Then the volume of a parallelopiped determined by vectors $\overline{OA}$, $\overline{OB}$, and $\overline{OC}$ (in cubic units), is

If $\bar{a}$ and $\bar{b}$ are unit vectors perpendicular to each other, then $\left[\bar{a} + (\bar{a} \times \bar{b})\quad \bar{b} + (\bar{a} \times \bar{b})\quad (\bar{a} \times \bar{b})\right] = \cdots$

If $\bar{a}$, $\bar{b}$ and $\bar{c}$ are three vectors such that $|\bar{a} + \bar{b} + \bar{c}| = 1$, $\bar{c} = \lambda(\bar{a} \times \bar{b})$ and $|\bar{a}| = \dfrac{1}{\sqrt{3}}$, $|\bar{b}| = \dfrac{1}{\sqrt{2}}$, $|\bar{c}| = \dfrac{1}{\sqrt{6}}$, then the angle between $\bar{a}$ and $\bar{b}$ is