The task is to compute the area of a triangle in a 3 dimensional space who is defined by those 3 points:
$$A=(1,1,1), B=(1,2,0), C=(2,1,2)$$
The problem is that the formula I know would compute me the volume of the triangle. The formula is the determinant of $([A][B][C])$ where $[V]$ represents V in the standard base. The area of a triangle is computed as half the area of a parallelogram.
Let’s use the cross product $\frac12|AB\times AC|$.
Note that the absolute value of the determinant of the three vectors gives the volume of the parallelepiped with sides OA OB OC.
$\vec {AB}=B-A=(0, 1, -1)$,$\,\vec {AC}=C-A=(1, 0, 1)$ so that $\vec {AB}\times\vec {AC}=(1, 1, -1)$. Thus the area of triangle is $\frac{1}{2}|\vec {AB}\times\vec {AC}|=\frac{\sqrt 3}{2}$.
