If the two diagonals of a parallelogram are $\bar{d_1} = \bar{i} + 2\bar{j} + 3\bar{k}$ and $\bar{d_2} = -2\bar{i} + \bar{j} - 2\bar{k}$,then the area of the parallelogram in square units is

$A$ tetrahedron has vertices $P(1, 2, 1)$,$Q(2, 1, 3)$,$R(-1, 1, 2)$ and $O(0, 0, 0)$. The angle between the faces $OPQ$ and $PQR$ is

Vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are such that $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$. $P_1$ and $P_2$ are two planes determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}, \vec{d}$ respectively. Then the angle between the planes $P_1$ and $P_2$ is

If $\theta$ is the angle between $\vec{f}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{g}=2 \hat{i}-3 \hat{j}+a \hat{k}$ and $\sin \theta=\sqrt{\frac{24}{28}}$,then $7 a^2+24 a=$

If $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$,$\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$,$|\overrightarrow{a}|=|\overrightarrow{b}|=2$,then $|\overrightarrow{u} \times \overrightarrow{v}|$ is equal to

If $a, b$ and $c$ are unit vectors such that $a+b+c=0$ and the angle between $a$ and $b$ is $\frac{\pi}{3}$, then $|a \times b|+|b \times c|+|c \times a|=$