$A$ tetrahedron has vertices at $O(0, 0, 0)$,$A(1, 2, 1)$,$B(2, 1, 3)$,and $C(-1, 1, 2)$. The angle between the faces $OAB$ and $ABC$ is:

If $a, b, c$ are vectors of equal magnitude such that $(a, b)=\alpha, (b, c)=\beta, (c, a)=\gamma$,then the minimum value of $\cos \alpha+\cos \beta+\cos \gamma$ is

If $\theta$ is the angle between the unit vectors $\mathbf{a}$ and $\mathbf{b}$,then $\cos \frac{\theta}{2} = $

Two adjacent sides of a parallelogram $PQRS$ are given by $\vec{PQ} = \hat{i} + \hat{k}$ and $\vec{PS} = \hat{i} - \hat{j}$. If the side $PS$ is rotated about the point $P$ by an acute angle $\alpha$ in the plane of the parallelogram so that it becomes perpendicular to the side $PQ$, then $\sin^2(\frac{5\alpha}{2}) - \sin^2(\frac{\alpha}{2})$ is equal to:

If $a$,$b$,$c$ are the $p^{th}$,$q^{th}$,$r^{th}$ terms of an $A.P.$ and $\vec x = (q - r)\hat i + (r - p)\hat j + (p - q)\hat k$ and $\vec y = a\hat i + b\hat j + c\hat k$,then:

If $\overrightarrow{a} \cdot \hat{i} = \overrightarrow{a} \cdot (\hat{i} + \hat{j}) = \overrightarrow{a} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$,then $\overrightarrow{a}$ is equal to