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

If the area of a parallelogram whose diagonals are represented by vectors $3 \hat{i}+\lambda \hat{j}+2 \hat{k}$ and $\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ is $\frac{\sqrt{117}}{2}$ sq. units, then $\lambda=$

If $\bar{a}, \bar{b}, \bar{c}$ are non coplanar unit vectors such that $\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{\sqrt{2}}$ then the angle between $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ is