If the lines $p_{1}x+q_{1}y=1,p_{2}x+q_{2}y=1$ and $p_{3}x+q_{3}y=1$ be concurrent, then the points $(p_1,q_1),(p_2,q_2)$ and $(p_3,q_3)$ ,

$p_{1}x+q_{1}y=1,p_{2}x+q_{2}y=1p_{3}x+q_{3}y=1$
Given lines are concurrent
$\Rightarrow \left|\begin{array}{ccc}{p}_1& q_1& 1\\ p_2& q_2& 1\\ p_3& q_3& 1\end{array}\right|=0$

$\Rightarrow p_1(q_2-q_3)-q_1(p_2-p_3)+(p_2{q}_3-p_3{q}_2)=0$

$\Rightarrow (p_1{q}_2-p_2{q}_1)+(p_2{q}_3-p_3{q}_2)+(p_3{q}_1-p_1{q}_3)=0$
The left hand side of the above equation is also equal to twice the area of a triangle with coordinates $(p_1,q_1),(p_2,q_2),(p_3,q_3)$
Since it is equal to zero, $(p_1,q_1),(p_2,q_2),(p_3,q_3)$ are collinear.