Answer
\[\sqrt {38} \]
Work Step by Step
\[\begin{gathered}
{\mathbf{u}} = - 3{\mathbf{i}} + 2{\mathbf{k}},\,\,\,{\mathbf{v}} = {\mathbf{i}} + {\mathbf{j}} + {\mathbf{k}} \hfill \\
{\text{The area of the parallelogram is given by }}\left| {{\mathbf{u}} \times {\mathbf{v}}} \right| \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{ - 3}&0&2 \\
1&1&1
\end{array}} \right| \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
0&2 \\
1&1
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{ - 3}&2 \\
1&1
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{ - 3}&0 \\
1&1
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = \left( { - 2} \right){\mathbf{i}} - \left( { - 3 - 2} \right){\mathbf{j}} + \left( { - 3} \right){\mathbf{k}} \hfill \\
{\mathbf{u}} \times {\mathbf{v}} = - 2{\mathbf{i}} + 5{\mathbf{j}} - 3{\mathbf{k}} \hfill \\
{\text{Area}} = \left| {{\mathbf{u}} \times {\mathbf{v}}} \right| \hfill \\
{\text{Area}} = \left| { - 2{\mathbf{i}} + 5{\mathbf{j}} - 3{\mathbf{k}}} \right| \hfill \\
{\text{Area}} = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( 5 \right)}^2} + {{\left( { - 3} \right)}^2}} \hfill \\
{\text{Area}} = \sqrt {4 + 25 + 9} \hfill \\
{\text{Area}} = \sqrt {38} \hfill \\
\end{gathered} \]
