The area of a triangle with vertices $(1, 2, 0)$,$(1, 0, a)$,and $(0, 3, 1)$ is $\sqrt{6}$ sq. units. Then the values of '$a$' are:

$A$ unit vector perpendicular to the vectors $4i - j + 3k$ and $-2i + j - 2k$ is

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

The unit vector perpendicular to both the vectors $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ is

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. If $\overline{c}$ is a vector such that $\overline{a} \cdot \overline{c}=|\overline{c}|$,$|\overline{c}-\overline{a}|=2 \sqrt{2}$ and the angle between $(\overline{a} \times \overline{b})$ and $\overline{c}$ is $\frac{\pi}{6}$,then $|(\overline{a} \times \overline{b}) \times \overline{c}|$ is

If the points $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$ are coplanar,find the unit vector perpendicular to the plane containing these points.