Let $\bar{a}=\hat{i}+\hat{j}-\hat{k}$ and $\bar{c}=5 \hat{i}-3 \hat{j}+2 \hat{k}$ and if $\overline{\mathrm{b}} \times \overline{\mathrm{c}}=\overline{\mathrm{a}}$ then $|\overline{\mathrm{b}}|=$

If $\bar{a}=\hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{j}-\hat{k}$ then a vector $\bar{c}$ such that $\overline{\mathrm{a}} \times \overline{\mathrm{c}}=\overline{\mathrm{b}}$ and $\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=3$ is

A tetrahedron has vertices $\mathrm{O}(0,0,0), \mathrm{A}(1,2,1)$, $B(2,1,3), C(-1,1,2)$. Then the angle between the faces OAB and ABC will be

The position vectors of the points $A, B, C$ are $\hat{i}+2 \hat{j}-\hat{k}, \hat{i}+\hat{j}+\hat{k}, 2 \hat{i}+3 \hat{j}+2 \hat{k}$ respectively. If $A$ is chosen as the origin, then the cross product of position vectors of $B$ and $C$ are