Let $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+2 \hat{k}$. Then the number of vectors $\vec{b}$ such that $\vec{b} \times \vec{c}=\vec{a}$ and $|\vec{b}| \in\{1, 2, \ldots, 10\}$ is

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 $p/q = \dots$

For $\vec{a}$ and $\vec{b}$,$|\vec{a}|=3$,$|\vec{b}|=\frac{\sqrt{2}}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector,then the angle between $\vec{a}$ and $\vec{b}$ is . . . . . . .

If $|\vec{a}| = 4$,$|\vec{b}| = 2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then $|\vec{a} \times \vec{b}|^2 = \dots$

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 :

Suppose $L_1$ and $L_2$ are two lines having the direction ratios $1, -2, -2$ and $0, 2, 1$ respectively. If the direction cosines of a line perpendicular to both $L_1$ and $L_2$ are $l, m, n$, then $|l| + |m| + |n| =$