Answer
$area = \sqrt {69} $
Work Step by Step
We have
$P = \left( {1,1,5} \right)$, ${\ \ }$ $Q = \left( {3,4,3} \right)$, ${\ \ }$ and ${\ \ }$ $R = \left( {1,5,7} \right)$
Write
${\bf{u}} = \overrightarrow {PR} = \left( {1,5,7} \right) - \left( {1,1,5} \right) = \left( {0,4,2} \right)$
and
${\bf{v}} = \overrightarrow {PQ} = \left( {3,4,3} \right) - \left( {1,1,5} \right) = \left( {2,3, - 2} \right)$
Using Eq. (7), the area of the triangle with vertices $P = \left( {1,1,5} \right)$, $Q = \left( {3,4,3} \right)$, and $R = \left( {1,5,7} \right)$ is
$area = \frac{{||{\bf{u}} \times {\bf{v}}||}}{2}$
First, we evaluate the vector product ${\bf{u}} \times {\bf{v}}$:
${\bf{u}} \times {\bf{v}} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
0&4&2\\
2&3&{ - 2}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
4&2\\
3&{ - 2}
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
0&2\\
2&{ - 2}
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
0&4\\
2&3
\end{array}} \right|{\bf{k}}$
${\bf{u}} \times {\bf{v}} = - 14{\bf{i}} + 4{\bf{j}} - 8{\bf{k}}$
So,
$area = \frac{{||{\bf{u}} \times {\bf{v}}||}}{2} = \frac{1}{2}\sqrt {{{\left( { - 14} \right)}^2} + {4^2} + {{\left( { - 8} \right)}^2}} = \sqrt {69} $
