I'm given the following parametrization of a parallelogram
$(x,y,z)=(1,1,1)+s\cdot(2,-1,-1)+t\cdot(-1,3,2),s\in[0,1],t\in[0,1]$
I'm now asked to determine the area of this parallelogram as well the unit normal vector in the middle of the parallelogram.
It's been a while since I've dealt with vectors, so any help would be highly appreciated!
Take $P=(1,1,1)$ a vertex of the parallelogram. The two adiacent vertices are $Q=(3,0,0)$ (for $s=1$ and $t=0$) and $R=(0,4,3)$ (for $s=0$ and $t=1$). So the two vectors parallel to the two sides of the parallelogram $PQ$ and $PR$ are: $\vec v=Q-P=(2,-1,-1)^T$ and $\vec u=R-P=(-1,3,2)^T$.
Now remeber that the oriented area of a parallelogram is given by the corss product of the vectors parallel to two adiacent sides, so the area is the magnitude of the formal determinant: $$ \mathbf{A}= \det \begin{bmatrix} \vec i & \vec i & \vec k\\ 2&-1&-1\\ -1&3&2 \end{bmatrix} $$
Normalizing this ''vector'' you have a unit vector orthogonal to the parallelogram.