Answer
\[3\sqrt {10} \]
Work Step by Step
\[\begin{gathered}
{\mathbf{u}} = 2{\mathbf{i}} - {\mathbf{j}} - 2{\mathbf{k}},\,\,\,{\mathbf{v}} = 3{\mathbf{i}} + 2{\mathbf{j}} - {\mathbf{k}} \hfill \\
{\text{The area of the parallelogram is given by }}\left| {{\mathbf{u}} \times {\mathbf{v}}} \right| \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
2&{ - 1}&{ - 2} \\
3&2&{ - 1}
\end{array}} \right| \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
{ - 1}&{ - 2} \\
2&{ - 1}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
2&{ - 2} \\
3&{ - 1}
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
2&{ - 1} \\
3&2
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left( {1 + 4} \right){\mathbf{i}} - \left( { - 2 + 6} \right){\mathbf{j}} + \left( {4 + 3} \right){\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = 5{\mathbf{i}} - 4{\mathbf{j}} + 7{\mathbf{k}} \hfill \\
{\text{Area}} = \left| {{\mathbf{u}} \times {\mathbf{v}}} \right| \hfill \\
{\text{Area}} = \left| {5{\mathbf{i}} - 4{\mathbf{j}} + 7{\mathbf{k}}} \right| \hfill \\
{\text{Area}} = \sqrt {{{\left( 5 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 7 \right)}^2}} \hfill \\
{\text{Area}} = \sqrt {25 + 16 + 49} \hfill \\
{\text{Area}} = \sqrt {90} \hfill \\
{\text{Area}} = 3\sqrt {10} \hfill \\
\end{gathered} \]