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

Let $\bar{a}=\hat{i}+\hat{j}, \bar{b}=2 \hat{i}-\hat{k}, \bar{c}=3 \hat{i}-\hat{j}+\hat{k}$, then vector $\overline{\mathrm{p}}$ satisfying $\overline{\mathrm{p}} \cdot \overline{\mathrm{a}}=0$ and $\overline{\mathrm{p}} \times \overline{\mathrm{b}}=\overline{\mathrm{c}} \times \overline{\mathrm{b}}$ is

If $\bar{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \bar{b}=\hat{i}-2 \hat{j}-2 \hat{k}, \bar{c}=-\hat{i}+4 \hat{j}+3 \hat{k}$ and if $\overline{\mathrm{d}}$ is vector perpendicular to both $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}, \overline{\mathrm{a}} \cdot \overline{\mathrm{d}}=18$, then $|\overline{\mathrm{a}} \times \overline{\mathrm{d}}|^2=$