Answer
$$11$$
Work Step by Step
\[\begin{gathered}
{\mathbf{u}} = 3{\mathbf{i}} - {\mathbf{j}},\,\,\,{\mathbf{v}} = 3{\mathbf{j}} + 2{\mathbf{k}} \hfill \\
{\text{The area of the parallelogram is given by }}\left| {{\mathbf{u}} \times {\mathbf{v}}} \right| \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
3&{ - 1}&0 \\
0&3&2
\end{array}} \right| \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
{ - 1}&0 \\
3&2
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
3&0 \\
0&2
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
3&{ - 1} \\
0&3
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left( { - 2 - 0} \right){\mathbf{i}} - \left( {6 - 0} \right){\mathbf{j}} + \left( {9 - 0} \right){\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = - 2{\mathbf{i}} - 6{\mathbf{j}} + 9{\mathbf{k}} \hfill \\
{\text{Area}} = \left| {{\mathbf{u}} \times {\mathbf{v}}} \right| \hfill \\
{\text{Area}} = \left| { - 2{\mathbf{i}} - 6{\mathbf{j}} + 9{\mathbf{k}}} \right| \hfill \\
{\text{Area}} = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( 9 \right)}^2}} \hfill \\
{\text{Area}} = \sqrt {121} \hfill \\
{\text{Area}} = 11 \hfill \\
\end{gathered} \]
