Solution:

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3} & 1\end{array}\right|$
$=\frac{1}{2}\left|\begin{array}{ccc}
-2 & 5 & 1 \\
-6 & -7 & 1 \\
-5 & -4 & 1
\end{array}\right|$
Expanding with $\mathrm{C}_{3}$
$\begin{array}{l}
=\frac{1}{2}[(24-35)-(8+25)+(14+30)]=\frac{1}{2}[-11-33+44] \\
=0
\end{array}$
Since the area between the 3 points is 0 , the three points lie in a straight line, i.e. they are collinear.