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

If non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular to each other,then the solution of the equation $\vec{r} \times \vec{a} = \vec{b}$ is:

$A$ vector of length $3$ perpendicular to each of the vectors $3\,i + j - 4\,k$ and $6\,i + 5\,j - 2\,k$ is

Let $\hat{a}$ be a unit vector perpendicular to the vectors $\overrightarrow{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\overrightarrow{c} = 2\hat{i} + 3\hat{j} - \hat{k}$,and makes an angle of $\cos^{-1}\left(-\frac{1}{3}\right)$ with the vector $\hat{i} + \hat{j} + \hat{k}$. If $\hat{a}$ makes an angle of $\frac{\pi}{3}$ with the vector $\hat{i} + \alpha\hat{j} + \hat{k}$,then the value of $\alpha$ is:

Let the vectors $\overline{PQ}, \overline{QR}, \overline{RS}, \overline{ST}, \overline{TU}$ and $\overline{UP}$ represent the sides of a regular hexagon.
$STATEMENT-1$: $\overline{PQ} \times (\overline{RS} + \overline{ST}) \neq \overrightarrow{0}$.
$STATEMENT-2$: $\overline{PQ} \times \overline{RS} = \overrightarrow{0}$ and $\overline{PQ} \times \overline{ST} \neq \overrightarrow{0}$.

Let $ABCD$ be a quadrilateral with $AB = a$, $AD = b$ and $AC = 3a + 2b$. If its area is $\alpha$ times the area of the parallelogram with $AB$ and $AD$ as adjacent sides, then the value of $\alpha$ is equal to