Find the unit vector perpendicular to the plane containing the points $(1, -1, 2), (2, 0, -1),$ and $(0, 2, 1)$.

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors such that $\overline{a} \neq \overline{0}$ and $\overline{a} \times \overline{b} = 2 \overline{a} \times \overline{c}$,$|\overline{a}| = |\overline{c}| = 1$,$|\overline{b}| = 4$ and $|\overline{b} \times \overline{c}| = \sqrt{15}$. If $\overline{b} - 2 \overline{c} = \lambda \overline{a}$,then $\lambda$ is

Find the unit vector perpendicular to both vectors $\vec{a}$ and $\vec{b}$.

Let $\overrightarrow{a}=3\hat{i}-\hat{j}+2\hat{k}$,$\overrightarrow{b}=\overrightarrow{a}\times(\hat{i}-2\hat{k})$ and $\overrightarrow{c}=\overrightarrow{b}\times\hat{k}$. Then the projection of $\overrightarrow{c}-2\hat{j}$ on $\overrightarrow{a}$ is:

The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8\hat{i} - 6\hat{j}$ and $3\hat{i} + 4\hat{j} - 12\hat{k}$ is:

Let $\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$ and $\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}$. Then the square of the area of the triangle with adjacent sides determined by the vectors $(2\vec{a} + 3\vec{b})$ and $(\vec{a} - \vec{b})$ is :