$A$ unit vector in the plane of $i + 2j + k$ and $i + j + 2k$ which is perpendicular to $2i + j + k$ is

If $|a| = 4$,$|b| = 2$ and the angle between $a$ and $b$ is $\pi/6$,then find $|a \times b|^{2}$.

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 $|(a \times b) \times c| = \dots$

For a triangle $ABC$, let $\vec{p}=\vec{BC}$, $\vec{q}=\vec{CA}$ and $\vec{r}=\vec{BA}$. If $|\vec{p}|=2\sqrt{3}$, $|\vec{q}|=2$ and $\cos \theta = \frac{1}{\sqrt{3}}$ where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$, then $|\vec{p} \times (\vec{q}-3\vec{r})|^{2}+3|\vec{r}|^{2}$ is equal to:

$A$ force $\overrightarrow{F} = 2i + j - k$ acts at a point $A$,whose position vector is $2i - j$. The moment of $\overrightarrow{F}$ about the origin is:

If the vertices of $\Delta ABC$ are $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$,then what is the area of the triangle?