Solution of equation $\frac{d y}{d x}=\frac{3 x-4 y-2}{3 x-4 y-3}$ is :

$x-y+c=\log (3 x-4 y+1)$

Let 3x – 4y = z

$3-4 \frac{d y}{d x}=\frac{d z}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{4}\left(3-\frac{d z}{d x}\right)$

Therefore the given equation $\frac{3}{4}-\frac{1}{4} \frac{d z}{d x}=\frac{z-2}{z-3}$

$\Rightarrow-\left(\frac{z-3}{z+1}\right) d z=d x \Rightarrow-\left(1-\frac{4}{z+1}\right) dz=dx$

$\Rightarrow-z+4 \log (z+1)=x+c \Rightarrow \log (3 x-4 y+1)=x-y+c$

Hence (2) is the correct answer.