Answer
$\sqrt{129}$
Work Step by Step
The first job is to find pairs of parallel vectors, so we are able to tell which two vectors that originate from a point (say, A) define the parallelogram.
The three vectors with initial point A are
$ \overrightarrow{AB} = \langle 3-0, 2-0,4-0 \rangle= \langle 3, 2, 4 \rangle$
$ \overrightarrow{AC} = \langle 5-0, 1-0,4-0 \rangle= \langle 5, 1, 4 \rangle$
$ \overrightarrow{AD} = \langle 2, -1, 0 \rangle$
We search for parallel vectors
$ \overrightarrow{BC} = \langle 5-3, 1-2,4-4 \rangle= \langle 2, -1, 0 \rangle,$ which is parallel to $ \overrightarrow{AD}$,
and has the same orientation.
$ \overrightarrow{CD} = \langle 2-5, -1-1,0-4 \rangle= \langle-3, -2, -4 \rangle,$
So, $ \overrightarrow{DC}$ is parallel to $ \overrightarrow{AB}$.
We have our pairs of parallel vectors.
The vectors defining the paralelogram are $\overrightarrow{AB}$ and $\overrightarrow{AD}$ ($\overrightarrow{AC}$ is the diagonal).
Area =$| \overrightarrow{AB}\times \overrightarrow{AD} |$
$\overrightarrow{AB}\times\overrightarrow{AD}$=$\left|\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k}\\
3 & 2 & 4\\
2 & -1 & 0
\end{array}\right|=\langle 0+4 ,\ -(0-8) ,\ -3-4 \rangle$
$=\langle 4 ,\ 8 ,\ -7 \rangle$
Area =$| \overrightarrow{AB}\times\overrightarrow{AD} |$=$\sqrt{16+64+49}=\sqrt{129}$