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.

(N/A) Let the vertices of the triangle be $A, B, C$ with position vectors $\vec{a}, \vec{b}, \vec{c}$ respectively.
The vector area of $\Delta ABC$ is given by $\frac{1}{2}(\overrightarrow{AB} \times \overrightarrow{AC})$.
We have $\overrightarrow{AB} = \vec{b} - \vec{a}$ and $\overrightarrow{AC} = \vec{c} - \vec{a}$.
Vector area $= \frac{1}{2}[(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})]$
$= \frac{1}{2}[\vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{a} \times \vec{c} + \vec{a} \times \vec{a}]$
Since $\vec{a} \times \vec{a} = 0$,$-\vec{b} \times \vec{a} = \vec{a} \times \vec{b}$,and $-\vec{a} \times \vec{c} = \vec{c} \times \vec{a}$,we get:
Vector area $= \frac{1}{2}[\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a}]$.
For the points to be collinear,the area of the triangle must be zero.
Therefore,$\frac{1}{2}[\vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b}] = 0$,which implies $\vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b} = 0$.
The unit vector $\hat{n}$ normal to the plane is given by $\hat{n} = \frac{\overrightarrow{AB} \times \overrightarrow{AC}}{|\overrightarrow{AB} \times \overrightarrow{AC}|} = \frac{\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}}{|\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}|}$.