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

\(\displaystyle{B}{C}={<}{5}+{1},{2}-{3}{>}\)

\(\displaystyle={<}{6},-{1}{>}\)

And, the other vector side BD is

\(\displaystyle{B}{D}={<}{3}+{1},-{1}-{3}{>}\)

\(\displaystyle={<}{4},-{4}{>}\)

And the cross product of both vectors is

\(\displaystyle{B}{C}\times{B}{D}={<}{6},-{1}{>}\times{<}{4},-{4}{>}\)

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

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

\(\displaystyle={0}{i}-{0}{j}+{k}{\left({\left({6}\right)}{\left(-{4}\right)}-{\left({1}\right)}{\left(-{4}\right)}\right)}\)

\(\displaystyle=-{20}{k}\)

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

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

\(\displaystyle={20}\)

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