The vector $x$ is perpendicular to the vectors $a=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $b=18 \hat{i}-22 \hat{j}-5 \hat{k}$,and it makes an obtuse angle with $\hat{j}$. If $|x|=14$,then $x=$

Let $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$. Let $\vec c$ be a vector such that $|\vec c - \vec a| = 3$,$|(\vec a \times \vec b) \times \vec c| = 3$,and the angle between $\vec c$ and $\vec a \times \vec b$ is $30^\circ$. Then $\vec a \cdot \vec c$ is equal to:

$A$ unit vector which is perpendicular to $i + 2j - 2k$ and $-i + 2j + 2k$ is

$\vec{a}, \vec{b}, \vec{c}$ are three vectors each having $\sqrt{2}$ magnitude such that $(\vec{a}, \vec{b})=(\vec{b}, \vec{c})=(\vec{c}, \vec{a})=\frac{\pi}{3}$. If $\vec{x}=\vec{a} \times(\vec{b} \times \vec{c})$ and $\vec{y}=\vec{b} \times(\vec{c} \times \vec{a})$, then

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 $\overline{PQ} \times \overline{ST} \neq \overrightarrow{0}$.

The area of the parallelogram whose diagonals are the vectors $2\vec{a} - \vec{b}$ and $4\vec{a} - 5\vec{b},$ where $\vec{a}$ and $\vec{b}$ are unit vectors forming an angle of $45^{\circ},$ is