Solution:
The given points are $A(1,-1,3), B(2,-4,5)$ and $(5,-13,11)$.
We need to prove collinear,
$\begin{array}{l}
\mathrm{AB}=\sqrt{(1-2)^{2}+(-1+4)^{2}+(3-5)^{2}}=\sqrt{1+9+4}=\sqrt{14} \\
\mathrm{BC}=\sqrt{(2-5)^{2}+(-4+13)^{2}+(5-11)^{2}}=\sqrt{9+81+36}=3 \sqrt{14} \\
\mathrm{AC}=\sqrt{(1-5)^{2}+(-1+13)^{2}+(3-11)^{2}}=\sqrt{16+144+64}=4 \sqrt{14} \\
\therefore \mathrm{AB}+\mathrm{BC}=\sqrt{14}+3 \sqrt{14} \\
=4 \sqrt{14} \\
=\mathrm{AC}
\end{array}$
$\therefore$ The points $A, B$ and $C$ are collinear.