$A$ vector perpendicular to both of the vectors $i + j + k$ and $i + j$ is

If the position vectors of the vertices of a $\triangle ABC$ are $\vec{OA} = 3\hat{i} + \hat{j} + 2\hat{k}$,$\vec{OB} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{OC} = 2\hat{i} + 3\hat{j} + \hat{k}$,then the length of the altitude of $\triangle ABC$ drawn from $A$ is

Let $a = 2i + j - 2k$ and $b = i + j$. If $c$ is a vector such that $a \cdot c = |c|$,$|c - a| = 2\sqrt{2}$,and the angle between $(a \times b)$ and $c$ is $30^\circ$,then $|(a \times b) \times c| = $

The vector $x\hat{i} + y\hat{j} + z\hat{k}$ makes an acute angle $\cot^{-1} \sqrt{2}$ with the plane containing the vectors $(2, 3, -1)$ and $(1, -1, 2)$. Then,

Vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are such that $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$. $P_1$ and $P_2$ are two planes determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}, \vec{d}$ respectively. Then the angle between the planes $P_1$ and $P_2$ is

The sine of the angle between the two vectors $3i + 2j - k$ and $12i + 5j - 5k$ will be