Answer
$area = 4\sqrt 2 $
Work Step by Step
Using Eq. (6), the parallelogram spanned by ${\bf{u}} = \left( {1,1,1} \right)$ and ${\bf{v}} = \left( {0,0,4} \right)$ is
$area = ||{\bf{u}} \times {\bf{v}}||$
First, we evaluate the vector product ${\bf{u}} \times {\bf{v}}$:
${\bf{u}} \times {\bf{v}} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&1&1\\
0&0&4
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
1&1\\
0&4
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
1&1\\
0&4
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
1&1\\
0&0
\end{array}} \right|{\bf{k}}$
${\bf{u}} \times {\bf{v}} = 4{\bf{i}} - 4{\bf{j}}$
So,
$area = ||{\bf{u}} \times {\bf{v}}|| = \sqrt {{4^2} + {{\left( { - 4} \right)}^2}} = \sqrt {32} = 4\sqrt 2 $
