For three lines to be concurrent, the determinant of their coefficients must be zero:
$\begin{vmatrix} a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1 \end{vmatrix} = 0$.
Apply elementary row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$\begin{vmatrix} a & 2 & 1 \\ b - a & 1 & 0 \\ c - b & 1 & 0 \end{vmatrix} = 0$.
Expanding along the third column: $1 \cdot ((b - a)(1) - (c - b)(1)) = 0$.
$b - a - c + b = 0 \implies 2b = a + c$.
Hence, $a, b, c$ are in Arithmetic Progression (A.P.).
Hence proved that $a, b, c$ are in A.P.