Solution:

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3} & 1\end{array}\right|$
Since they are collinear, the area will be 0
$\rightarrow 0=\frac{1}{2}\left|\begin{array}{ccc}
5 & 5 & 1 \\
k & 1 & 1 \\
11 & 7 & 1
\end{array}\right|$
Expanding with $C_{3}$
$\begin{array}{l}
\rightarrow 0=(7 \mathrm{k}-11)-(35-55)+(5-5 \mathrm{k}) \\
\rightarrow 0=2 \mathrm{k}-14 \\
\rightarrow 2 \mathrm{k}=14 \\
\therefore \mathrm{k}=7
\end{array}$