Answer
$area = \frac{9}{2}\sqrt 3 $
Work Step by Step
Write ${\bf{u}} = \overrightarrow {OP} = \left( {3,3,0} \right)$ and ${\bf{v}} = \overrightarrow {OQ} = \left( {0,3,3} \right)$.
Using Eq. (7) the triangle with vertices at the origin $O$, $P = \left( {3,3,0} \right)$, and $Q = \left( {0,3,3} \right)$ is
$area = \frac{{||{\bf{u}} \times {\bf{v}}||}}{2}$
First, we evaluate the vector product ${\bf{u}} \times {\bf{v}}$:
${\bf{u}} \times {\bf{v}} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
3&3&0\\
0&3&3
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
3&0\\
3&3
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
3&0\\
0&3
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
3&3\\
0&3
\end{array}} \right|{\bf{k}}$
${\bf{u}} \times {\bf{v}} = 9{\bf{i}} - 9{\bf{j}} + 9{\bf{k}}$
So,
$area = \frac{{||{\bf{u}} \times {\bf{v}}||}}{2} = \frac{1}{2}\sqrt {{9^2} + {{\left( { - 9} \right)}^2} + {9^2}} = \frac{9}{2}\sqrt 3 $
