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

Let the vectors $\overline{PQ}, \overline{QR}, \overline{RS}, \overline{ST}, \overline{TU}$ and $\overline{UP}$ represent the sides of a regular hexagon.
$STATEMENT-1$: $\overline{PQ} \times (\overline{RS} + \overline{ST}) \neq \overrightarrow{0}$.
$STATEMENT-2$: $\overline{PQ} \times \overline{RS} = \overrightarrow{0}$ and $\overline{PQ} \times \overline{ST} \neq \overrightarrow{0}$.

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$,$\overrightarrow{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$,$\overrightarrow{c} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\overrightarrow{d}$ be a vector such that $\overrightarrow{b} \times \overrightarrow{d} = \overrightarrow{c} \times \overrightarrow{d}$ and $\overrightarrow{a} \cdot \overrightarrow{d} = 4$. Then $|(\overrightarrow{a} \times \overrightarrow{d})|^2$ is equal to . . . . . . .

If $\vec{u} = \vec{a} - \vec{b}$ and $\vec{v} = \vec{a} + \vec{b}$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\vec{u} \times \vec{v}| = ......$

If $|\vec{a}|=10, |\vec{b}|=2$ and $\vec{a} \cdot \vec{b}=12$,then $|\vec{a} \times \vec{b}|=$ . . . . . . .

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k},$ find a vector $\vec{c}$ such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3.$