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

Find $|a \times b|^2$,if $|a|=2, |b|=3$ and the angle between $a$ and $b$ is $\theta = \frac{\pi}{6}$.

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vertices of a triangle,show that $\frac{1}{2}[\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}]$ gives the vector area of the triangle. Hence,deduce the condition that the three points $\vec{a}, \vec{b}, \vec{c}$ are collinear. Also,find the unit vector normal to the plane of the triangle.

Let the vectors $\overline{PQ}, \overline{QR}, \overline{RS}, \overline{ST}, \overline{TU}$ and $\overline{UP}$ represent the sides of a regular hexagon.
$STATEMENT-1$: $\overline{PQ} \times (\overline{RS} + \overline{ST}) \neq \overrightarrow{0}$.
$STATEMENT-2$: $\overline{PQ} \times \overline{RS} = \overrightarrow{0}$ and $\overline{PQ} \times \overline{ST} \neq \overrightarrow{0}$.

If $\bar{a}=2 \hat{i}+3 \hat{j}-\hat{k}, \bar{b}=-\hat{i}+2 \hat{j}-4 \hat{k}$ and $\bar{c}=\hat{i}+\hat{j}-2 \hat{k}$,then $(\bar{a} \times \bar{b}) \cdot(\bar{a} \times \bar{c})=$

Let $a$,$b$,and $c$ be real numbers such that $a^2 + b^2 + c^2 = 1$. Then the maximum value of $(4b - 3c)^2 + (4a - 2c)^2 + (3a - 2b)^2$ is: