The adjacent sides of a parallelogram are $\bar{a} = \hat{j} + 2\hat{k}$ and $\bar{b} = \hat{i} + 2\hat{j}$. Find its area.

$A, B, C, D$ are any $4$ points and $|\overline{AB} \times \overline{CD} + \overline{BC} \times \overline{AD} + \overline{CA} \times \overline{BD}| = \lambda$ (Area of $\triangle ABC$). Then $\lambda = $

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}$ and $\hat{i}+\hat{j}$ and the plane determined by vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$. The angle between $\vec{a}$ and $(\hat{i}-2\hat{j}+2\hat{k})$ is

The magnitude of the projection of the vector $2\hat{i}+\hat{j}+\hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+3\hat{k}$ is:

$A$ vector with magnitude of $3$ units,which is perpendicular to each of the vectors $\vec{a}=3 \hat{i}+\hat{j}-4 \hat{k}$ and $\vec{b}=6 \hat{i}+5 \hat{j}-2 \hat{k}$,is given by

$A$ unit vector perpendicular to the plane of $a = 2i - 6j - 3k$ and $b = 4i + 3j - k$ is