Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ and $-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ respectively is

$A$ tetrahedron has vertices $O(0,0,0)$,$A(1,2,1)$,$B(2,1,3)$,and $C(-1,1,2)$. The angle between the faces $OAB$ and $ABC$ is:

The area of the parallelogram whose diagonals are $\frac{3}{2}i + \frac{1}{2}j - k$ and $2i - 6j + 8k$ is

Consider the cube in the first octant with sides $OP, OQ$ and $OR$ of length $1$,along the $x$-axis,$y$-axis and $z$-axis,respectively,where $O(0,0,0)$ is the origin. Let $S\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ be the centre of the cube and $T$ be the vertex of the cube opposite to the origin $O$ such that $S$ lies on the diagonal $OT$. If $\overrightarrow{p} = \overrightarrow{SP}, \overrightarrow{q} = \overrightarrow{SQ}, \overrightarrow{r} = \overrightarrow{SR}$ and $\overrightarrow{t} = \overrightarrow{ST}$,then the value of $|(\overrightarrow{p} \times \overrightarrow{q}) \times (\overrightarrow{r} \times \overrightarrow{t})|$ is:

Let $a=2 \hat{i}-2 \hat{j}+\hat{k}$ and $b=-\hat{j}+\hat{k}$. If $c$ is a vector such that $a \cdot c=|c|$,$|c-a|=2 \sqrt{2}$,and the angle between $a \times b$ and $c$ is $\frac{\pi}{3}$,then $|(a \times b) \times c|=$

$A, B, C, D$ 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 = $