Answer
\[4\sqrt 2 \]
Work Step by Step
\[\begin{gathered}
A\left( {5,6,2} \right),\,\,B\left( {7,16,4} \right),\,\,C\left( {6,7,3} \right) \hfill \\
\overrightarrow {AB} = \left\langle {2,10,2} \right\rangle \hfill \\
\overrightarrow {AC} = \left\langle {1,1,1} \right\rangle \hfill \\
{\text{The area of the triangle is given by }}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
2&{10}&2 \\
1&1&1
\end{array}} \right| \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}
{10}&2 \\
1&1
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
2&2 \\
1&1
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
2&{10} \\
1&1
\end{array}} \right|{\mathbf{k}} \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = \left( {10 - 2} \right){\mathbf{i}} - \left( 0 \right){\mathbf{j}} + \left( {2 - 10} \right){\mathbf{k}} \hfill \\
\overrightarrow {AB} \times \overrightarrow {AC} = 8{\mathbf{i}} + 0{\mathbf{j}} - 8{\mathbf{k}} \hfill \\
\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {{{\left( 8 \right)}^2} + {{\left( 0 \right)}^2} + {{\left( { - 8} \right)}^2}} \hfill \\
\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {128} \hfill \\
\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = 8\sqrt 2 \hfill \\
{\text{The triangle with vertices }}A{\text{, }}B{\text{, and }}C{\text{ comprises }} \hfill \\
{\text{half of the parallelogram}}{\text{, so its area is}} \hfill \\
{\text{Area}} = 4\sqrt 2 \hfill \\
\end{gathered} \]
