Calculations:

From the given points, we may choose any three of then such and dind a two vector that express the sides of the parallelogram "or one side and a diagonal of the parallelogram", we may chose for example the points B, C, and D.

More over, we can choose point B to be the common point for the two sides "the initial point of the two vector", thus we need to find vector BD and vector BC, using equation as follows

\(BC \leq 5+1,2-3>\)

\(\leq 6,-1>\)

And, the other vector side BD is

\(BD \leq 3+1,-1-3>\)

\( \leq 4,-4>\)

And the cross product of both vectors is

\(BC\times BD \leq 6,-1>\times<4,-4>\)

\(=\begin{vmatrix}i&j&k\\6&-1&0\\4&-4&0\end{vmatrix}\)

\(=i\begin{vmatrix}-1&0\\4&0\end{vmatrix}-j\begin{vmatrix}6&0\\4&0\end{vmatrix}+k\begin{vmatrix}6&-1\\4&-4\end{vmatrix}\)

\(=0i-0j+k((6)(-4)-(1)(-4))\)

\(=-20k\)

Knowing, the cross product of the two vectors of the parallelogram we can use equation to find the area

\(\text{Area}=|-20k|\)

\(=20\)

Thus, the area of the parallelogram is 20 units squared.