Answer
$2\sqrt {138}$
Work Step by Step
The area of the parallelogram= $||\textbf{v}\times\textbf{w}||$
We calculate the cross product as follows:
$\textbf{v}\times\textbf{w}=\begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\1&3&1\\-4&2&6\end{vmatrix}$
$=\textbf{i}(6\times3-2\times1)-\textbf{j}(6\times1-(-4\times1))+\textbf{k}(2\times1-(3\times-4))$
$=16\textbf{i}-10\textbf{j}+14\textbf{k}$
$||\textbf{v}\times\textbf{w}||=\sqrt {16^{2}+10^{2}+14^{2}}$
$=2\sqrt {138}$
