Answer
\[\frac{{\sqrt {33} }}{2}\]
Work Step by Step
\[\begin{gathered}
A\left( {1,2,3} \right),\,\,B\left( {5,1,5} \right),\,\,C\left( {2,3,3} \right) \hfill \\
\overrightarrow {AB} = \left\langle {4, - 1,2} \right\rangle \hfill \\
\overrightarrow {AC} = \left\langle {1,1,0} \right\rangle \hfill \\
{\text{The area of the triangle is given by }}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
4&{ - 1}&2 \\
1&1&0
\end{array}} \right| \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}
{ - 1}&2 \\
1&0
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
4&2 \\
1&0
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
4&{ - 1} \\
1&1
\end{array}} \right|{\mathbf{k}} \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = \left( { - 2} \right){\mathbf{i}} - \left( { - 2} \right){\mathbf{j}} + \left( 5 \right){\mathbf{k}} \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = - 2{\mathbf{i}} + 2{\mathbf{j}} + 5{\mathbf{k}} \hfill \\
\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 5 \right)}^2}} \hfill \\
\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {33} \hfill \\
{\text{The triangle with vertices }}A{\text{, }}B{\text{, and }}C{\text{ comprises }} \hfill \\
{\text{half of the parallelogram}}{\text{, so its area is}} \hfill \\
{\text{Area}} = \frac{{\sqrt {33} }}{2} \hfill \\
\end{gathered} \]
