Consider the lines $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$. Then the unit vector perpendicular to both $L_1$ and $L_2$ is:

If $\overline{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\overline{OB} = 4\hat{i} + \hat{k}$ are the position vectors of the points $A$ and $B$,then the position vector of a point on the line passing through $B$ and parallel to the vector $\overline{OA} \times \overline{OB}$ which is at a distance of $\sqrt{189}$ units from $B$ is

Let $\overrightarrow{a}=3\hat{i}-\hat{j}+2\hat{k}$,$\overrightarrow{b}=\overrightarrow{a}\times(\hat{i}-2\hat{k})$ and $\overrightarrow{c}=\overrightarrow{b}\times\hat{k}$. Then the projection of $\overrightarrow{c}-2\hat{j}$ on $\overrightarrow{a}$ is:

If $a \times (b \times c) = 0,$ then

Let $\overrightarrow{a}$ be a non-zero vector parallel to the line of intersection of the two planes passing through the origin and containing the vectors $(\hat{i}+\hat{j}, \hat{i}+\hat{k})$ and $(\hat{i}-\hat{j}, \hat{j}-\hat{k})$ respectively. If $\theta$ is the angle between the vector $\vec{a}$ and the vector $\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$ and $\vec{a} \cdot \vec{b}=6$,then the ordered pair $(\theta, |\vec{a} \times \vec{b}|)$ is equal to

Let $\bar{u}=\hat{i}+\hat{j}$,$\bar{v}=\hat{i}-\hat{j}$ and $\bar{w}=\hat{i}+2\hat{j}+3\hat{k}$. If $\hat{n}$ is a unit vector such that $\bar{u} \cdot \hat{n}=0$ and $\bar{v} \cdot \hat{n}=0$,then $|\bar{w} \cdot \hat{n}|$ is equal to