How to calculate the area of a triangle ABC when given three position vectors $a, b$, and $ c$ in 3D? Where $a = ( 1, 2, 3), b = (2, 1, 3) $, and $c = (3,1,2). $

Cross product works great as a black box, but it lacks geometric intuition.

For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5\cdot \left \| \overrightarrow{AB} \right \|\cdot \left \| \overrightarrow{CH} \right \|$
$$\left \| \overrightarrow{CH} \right \|=\left \| \overrightarrow{AC} \right \| \cdot sin(θ)$$
where
$$θ = arccos(\frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{\left \| \overrightarrow{AB} \right \| \cdot \left \| \overrightarrow{AC} \right \|})$$
Thus:
$$Area=0.5\cdot \left \| \overrightarrow{AB} \right \|\cdot \left \| \overrightarrow{AC} \right \|\cdot sin(θ)$$
This is equal to:
$$0.5\cdot \left \| \overrightarrow{AB} \times  \overrightarrow{AC} \right \|$$
This is why 
$$0.5\cdot \left \| \overrightarrow{AB} \times  \overrightarrow{AC} \right \|=0.5\cdot \left \| \overrightarrow{AB} \right \|\cdot \left \| \overrightarrow{AC} \right \|\cdot sin(θ)$$
