Answer
\[\frac{{\sqrt {11} }}{2}\]
Work Step by Step
\[\begin{gathered}
A\left( {0,0,0,} \right),\,\,B\left( {3,0,1} \right),\,\,C\left( {1,1,0} \right) \hfill \\
{\text{The area of the triangle is given by }}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
3&0&1 \\
1&1&0
\end{array}} \right| \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}
0&1 \\
1&0
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
3&1 \\
1&0
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
3&0 \\
1&1
\end{array}} \right|{\mathbf{k}} \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = \left( { - 1} \right){\mathbf{i}} - \left( { - 1} \right){\mathbf{j}} + \left( 3 \right){\mathbf{k}} \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = - {\mathbf{i}} + {\mathbf{j}} + 3{\mathbf{k}} \hfill \\
\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 3 \right)}^2}} \hfill \\
\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {11} \hfill \\
{\text{The triangle with vertices }}A{\text{, }}B{\text{, and }}C{\text{ comprises }} \hfill \\
{\text{half of the parallelogram}}{\text{, so its area is}} \hfill \\
{\text{Area}} = \frac{{\sqrt {11} }}{2} \hfill \\
\end{gathered} \]