If $a = i + 2j - 2k$,$b = 2i - j + k$ and $c = i + 3j - k$,then $a \times (b \times c)$ is equal to

Let $\overline{OA} = \vec{a}$,$\overline{OB} = 10\vec{a} + 2\vec{b}$,and $\overline{OC} = \vec{b}$,where $O, A, C$ are non-collinear. Let $p$ be the area of the quadrilateral $OABC$ and $q$ be the area of the parallelogram with adjacent sides $OA$ and $OC$. Then $p/q = \dots$

Let $\overrightarrow{a}=2 \hat{i}-\hat{j}+3 \hat{k}$,$\overrightarrow{b}=3 \hat{i}-5 \hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c}=\vec{c} \times \vec{b}$ and $(\overrightarrow{a}+\overrightarrow{c}) \cdot(\overrightarrow{b}+\overrightarrow{c})=168$. Then the maximum value of $|\vec{c}|^2$ is :

Let $\vec{a}=\hat{i}+\alpha \hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}-\alpha \hat{j}+\hat{k}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $8 \sqrt{3}$ square units,then $\vec{a} \cdot \vec{b}$ is equal to ....... .

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

If $2 \hat{i}+3 \hat{j}-4 \hat{k}$ and $-\hat{i}+2 \hat{j}+\hat{k}$ are the two diagonals of a parallelogram, then the area of the parallelogram in square units is