Equation of line passing through point of intersection
of $L=0$ and $l=0$ is $L+\lambda l=0$
$$
\begin{aligned}
& \Rightarrow(2 x+3 y-5)+\lambda(4 x-5 y+7)=0 \\
& L_1:(2+4 \lambda) x+(3-5 \lambda) y+(7 \lambda-5)=0
\end{aligned}
$$
$\because L_1$ divides the line joining $(2,3)$ and $(1,-1)$ in $2: 1$


$\begin{aligned} & x=\frac{2+2}{3}=\frac{4}{3} ; y=\frac{-2+3}{3}=\frac{1}{3} \\ & \because \quad(x, y) \text { lies on } L_1 . \\ & \therefore \quad(2+4 \lambda) \cdot \frac{4}{3}+(3-5 \lambda) \cdot \frac{1}{3}+(7 \lambda-5)=0 \\ & \Rightarrow \quad 16 \lambda+8+3-5 \lambda+21 \lambda-15=0 \Rightarrow \lambda=\frac{1}{8} . \\ & \therefore \quad L_1:\left(2+\frac{4}{8}\right) x+\left(3-\frac{5}{8}\right) y+\left(\frac{7}{8}-5\right)=0 \\ & 20 x+19 y-33=0 \Rightarrow 20 x+19 y=33 \\ & \left(\frac{20}{33}\right) x+\left(\frac{19}{33}\right) y=1 \\ & \therefore \quad a=\frac{20}{33}, b=\frac{19}{33} \Rightarrow \quad 33(a-b)=33\left(\frac{20}{33}-\frac{19}{33}\right)=1 .\end{aligned}$