Answer
$5\sqrt 2$
Work Step by Step
Let the vector connecting the points (1,3,-1) and (2,-1,3) be $\textbf{v}$ and the vector connecting points (2,-1,3) and (4,1,1) be $\textbf{w}$. Then,
$\textbf{v}=(2-1)\textbf{i}+(-1-3)\textbf{j}+(3-(-1))\textbf{k}=\textbf{i}-4\textbf{j}+4\textbf{k}$
$\textbf{w}=(4-2)\textbf{i}+(1-(-1))\textbf{j}+(1-3)\textbf{k}= 2\textbf{i}+2\textbf{j}-2\textbf{k}$
The area of the triangle is given by:
$\frac{||\textbf{v}\times\textbf{w}||}{2}$
$\textbf{v}\times\textbf{w}=\begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\1&-4&4\\2&2&-2\end{vmatrix}$
$=\textbf{i}(-4\times-2-2\times4)-\textbf{j}(1\times-2-4\times2)+\textbf{k}(1\times2-(-4\times2))=10\textbf{j}+10\textbf{k}$
$||\textbf{v}\times\textbf{w}||=\sqrt {10^{2}+10^{2}}=\sqrt {200}=2\sqrt {50}$
Area= $\frac{||\textbf{v}\times\textbf{w}||}{2}=\sqrt {50}=5\sqrt 2$
