The line passing through the point $i + 3j + 2k$ and perpendicular to the lines $r = (i + 2j - k) + \lambda (2i + j + k)$ and $r = (2i + 6j + k) + \mu (i + 2j + 3k)$ is:

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}+\hat{k}$ are two vectors,and $\vec{c}$ is a unit vector lying in the plane of $\vec{a}$ and $\vec{b}$ such that $\vec{c}$ is perpendicular to $\vec{b}$,then find the value of $\vec{c} \cdot (\hat{i}+\hat{j}+2\hat{k})$.

$A$ vector $\vec{a}$ of length $2$ units makes an angle $60^{\circ}$ with each of the $X$-axis and $Y$-axis. If another vector $\vec{b}$ of length $\sqrt{2}$ units makes an angle $45^{\circ}$ with each of the $Y$-axis and $Z$-axis, then $\vec{a} \times \vec{b} = $

Let $\vec{a}, \vec{b},$ and $\vec{c}$ be three unit vectors,out of which vectors $\vec{b}$ and $\vec{c}$ are non-parallel. If $\alpha$ and $\beta$ are the angles which vector $\vec{a}$ makes with vectors $\vec{b}$ and $\vec{c}$ respectively and $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2} \vec{b},$ then $|\alpha - \beta|$ is equal to .............. $^o$

Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be the position vectors of the vertices $A, B, C$ respectively of $\triangle ABC$. The vector area of $\triangle ABC$ is:

For two given vectors $\bar{a}$ and $\bar{b}$,if the vectors $\overline{A}$ and $\overline{B}$ are such that $\overline{A}+\overline{B}=\bar{a}$,$\overline{A} \times \overline{B}=\bar{b}$,and $\overline{A} \cdot \bar{a}=1$,then $\overline{A}=$