$A$ 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:

Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a}+\vec{b}$ is a unit vector if $\theta =$

If $P(6, 10, 10)$,$Q(1, 0, -5)$,$R(6, -10, \lambda)$ are vertices of a triangle right-angled at $Q$,then the value of $\lambda$ is ....

The value of $b$ such that the scalar product of the vector $(i + j + k)$ with the unit vector parallel to the sum of the vectors $(2i + 4j - 5k)$ and $(bi + 2j + 3k)$ is $1$,is:

For $p > 0$,a vector $\vec{v}_{2} = 2 \hat{i} + (p + 1) \hat{j}$ is obtained by rotating the vector $\vec{v}_{1} = \sqrt{3} p \hat{i} + \hat{j}$ by an angle $\theta$ about the origin in the counter-clockwise direction. If $\tan \theta = \frac{(\alpha \sqrt{3} - 2)}{4 \sqrt{3} + 3}$,then the value of $\alpha$ is equal to $....$

In a parallelogram $OACB$,$\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and the foot of the perpendicular drawn from point $B$ to $AC$ is $M$. If $\vec{a} \cdot \vec{b} = 1$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\overrightarrow{BM}|$ is: