The position vectors of the points $A, B, C$ are $\hat{i}+2\hat{j}-\hat{k}, \hat{i}+\hat{j}+\hat{k}$,and $2\hat{i}+3\hat{j}+2\hat{k}$ respectively. If $A$ is chosen as the origin,then the cross product of the position vectors of $B$ and $C$ 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 $\frac{2 \pi}{3}$,then $|(\overline{a} \times \overline{b}) \times \overline{c}|=$

The direction cosines of a line which is perpendicular to lines whose direction ratios are $3, -2, 4$ and $1, 3, -2$ are

For vectors $\vec{a}$ and $\vec{b}$,if $|\vec{a}|=3$,$|\vec{b}|=\frac{\sqrt{2}}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector,then the angle between the two vectors $\vec{a}$ and $\vec{b}$ is . . . . . . .

$a$ and $b$ are two vectors such that $|a|=\sqrt{3}$ and $|b|=\sqrt{2}$. If $x$ is a unit vector satisfying $x \times a = b$,then $x$ is equal to:

Consider $P(1, 2, -3)$,$Q(-2, 1, -4)$,and $R(3, 4, -2)$. Let $\vec{B} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$,where $A_x, A_y$,and $A_z$ are the projections of the area of triangle $PQR$ on the $yz, zx$,and $xy$ planes,respectively. Then,the value of $|\vec{B}|^2$ is: