If  3a + 2b + 4c = 0 then the lines ax + by + c = 0 pass through the fixed point

$If3a+2b+4c=0\Rightarrow c=\frac{-3a}{4}-\frac{2b}{4}=\frac{-3a}{4}-\frac{b}{2}$ 

Given line ax + by + c = 0

$\begin{array}{l}\Rightarrow ax+by-\frac{3a}{4}-\frac{b}{2}=0\\ \Rightarrow a\left(x-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)+b\left(y-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=0\\ \Rightarrow \left(x-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)+\frac{b}{a}\left(y-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=0\\ It isoftheform{L}_1+\lambda L_2=0\\ L_1=x-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.=0,L_2=y-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=0\\ \therefore x=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,y=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ \therefore Thelineax+by+c=0passesthrough\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\end{array}$