Step 1 :

The given lines are:

$ax+2y+1=0 ; bx+3y+1=0 ; cx+4y+1=0$

It is said that the three lines are coplanar.

$ \therefore \begin{vmatrix} a & b & c \\ 2 & 3 & 4 \\ 1 & 1 & 1 \end{vmatrix}=0$

On expanding we get,

$a(3-4)-b(2-4)+c(2-3)=0$

$ \Rightarrow -a+2b-c=0$

$ \Rightarrow 2b=a+c$

This implies that a,b,c are in A.P.

Hence the statement is false.