Answer
$\displaystyle \frac{3\sqrt{2}}{2}$
Work Step by Step
$ \overrightarrow{AB} = \langle-1-0, 1-0,-1-0 \rangle= \langle-1, 1, -1 \rangle$
$ \overrightarrow{AC} = \langle 3, 0, 3 \rangle$
We find the area of the parallelogram defined by these two vectors.
The triangle has half the area of the parallelogram.
$A=\displaystyle \frac{1}{2}| \overrightarrow{AB} \times \overrightarrow{AC} |$
$ \overrightarrow{AB} \times \overrightarrow{AC}=\left|\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k}\\
-1 & 1 & -1\\
3 & 0 & 3
\end{array}\right|=\langle (3-0) ,\ -(-3+3) ,\ (0-3) \rangle$
$=\langle 3 ,\ 0 ,\ -3 \rangle$
$A=\displaystyle \frac{1}{2}\cdot\sqrt{9+0+9}=\frac{3\sqrt{2}}{2}$
