Area of triangle with vertices $(a_1, b_1), (a_2, b_2) \text{ and } (a_3, b_3)$ is

$\Delta = \frac{1}{2}\begin{vmatrix} a_1 & b_1 & 1 \\ a_2 & b_2 & 1 \\ a_3 & b_3 & 1 \end{vmatrix}$

As the three lines $a_1x + b_1y + c = 0$ , $a_2x + b_2y + c = 0$ and $a_3x + b_3y + c = 0$ are concurrent, so

$\begin{vmatrix} a_1 & b_1 & c \\ a_2 & b_2 & c \\ a_3 & b_3 & c \end{vmatrix} = 0$

$c\begin{vmatrix} a_1 & b_1 & 1 \\ a_2 & b_2 & 1 \\ a_3 & b_3 & 1 \end{vmatrix} = 0$

$c × \Delta = 0$

$\text{As } c \neq 0, \Delta = 0$

$\color{#3D99F6}{\boxed{\text{Area of triangle } = \frac{1}{2}\Delta = 0}}$

Let the point of concurrency be $(p, q)$ . Then $a_1p + b_1q + c = 0$ , $a_2p + b_2q + c = 0$ , and $a_3p + b_3q + c = 0$ .

Put another way, $px + qy + c = 0$ , where some solutions for $(x, y)$ are $(a_1, b_1)$ , $(a_2, b_2)$ , and $(a_3, b_3)$ .

Since $px + qy + c = 0$ is an equation of a line, $(a_1, b_1)$ , $(a_2, b_2)$ , and $(a_3, b_3)$ are collinear, so the triangle formed by these three points is degenerate and has an area of $\boxed{0}$ .