Answer
\[\frac{9}{2}\]
Work Step by Step
\[\begin{gathered}
A\left( { - 1, - 5 - 3} \right),\,\,B\left( { - 3, - 2, - 1} \right),\,\,C\left( {0, - 5, - 1} \right) \hfill \\
\overrightarrow {AB} = \left\langle { - 2,3,2} \right\rangle \hfill \\
\overrightarrow {AC} = \left\langle {1,0,2} \right\rangle \hfill \\
{\text{The area of the triangle is given by }}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{ - 2}&3&2 \\
1&0&2
\end{array}} \right| \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}
3&2 \\
0&2
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{ - 2}&2 \\
1&2
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{ - 2}&3 \\
1&0
\end{array}} \right|{\mathbf{k}} \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = \left( {6 - 0} \right){\mathbf{i}} - \left( { - 4 - 2} \right){\mathbf{j}} + \left( {0 - 3} \right){\mathbf{k}} \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = 6{\mathbf{i}} + 6{\mathbf{j}} - 3{\mathbf{k}} \hfill \\
\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {{{\left( 6 \right)}^2} + {{\left( 6 \right)}^2} + {{\left( { - 3} \right)}^2}} \hfill \\
\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {81} \hfill \\
\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = 9 \hfill \\
{\text{The triangle with vertices }}A{\text{, }}B{\text{, and }}C{\text{ comprises }} \hfill \\
{\text{half of the parallelogram}}{\text{, so its area is}} \hfill \\
{\text{Area}} = \frac{9}{2} \hfill \\
\end{gathered} \]
