Given:
\[{{p}_{1}}x\text{ }+\text{ }{{q}_{1}}y\text{ }=\text{ }1\]
\[{{p}_{2}}x\text{ }+\text{ }{{q}_{2}}y\text{ }=\text{ }1\]
and ,
\[{{p}_{3}}x\text{ }+\text{ }{{q}_{3}}y\text{ }=\text{ }1\]
The given lines can be written as follows:
\[{{p}_{1}}~x\text{ }+\text{ }{{q}_{1}}~y\text{ }\text{ }1\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 1 \right)\]
\[{{p}_{2}}~x\text{ }+\text{ }{{q}_{2}}~y\text{ }\text{ }1\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 2 \right)\]
and,
\[{{p}_{3}}~x\text{ }+\text{ }{{q}_{3}}~y\text{ }\text{ }1\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 3 \right)\]
since, the three lines are concurrent.
Now, consider the following determinant:
Hence proved, the given three points,
(p1, q1), (p2, q2) and (p3, q3) are collinear.