Equation of two path $\mathrm{OA}$ and $\mathrm{OB}$ are
$\begin{aligned}
&2 x-3 y+4=0 \\
&3 x+4 y-5=0
\end{aligned}$


on solving of both equations, we get
$\therefore x=-\frac{1}{17}, y=\frac{22}{17}$
$\therefore$ These path meet at $\left(-\frac{1}{17}, \frac{22}{17}\right)$
The third path $\mathrm{AB}$ whose eqn. is $6 x-7 y+8=0$.
The shortest path from $O$ to path $A B$ is $\perp$ to path $A B$.
$\therefore$ Slope of $\mathrm{AB}=\frac{6}{7}$
$\Rightarrow$ Slope of $\perp$ path $=\frac{-7}{6}$
$\therefore$ Equation of $\perp$ path $O M$ is,
$y-\frac{22}{17}=-\frac{7}{6}\left(x+\frac{1}{17}\right)$
$\begin{aligned}
&\Rightarrow \quad 17 y-22=-\frac{7}{6}(17 x+1) \\
&\Rightarrow \quad 102 y-132=-119 x-7 \\
&\Rightarrow \quad 119 x+102 y=132-7
\end{aligned}$
$\therefore \quad$ Equation of the shortest path is
$119 x+102 y-125=0$
Hence, the time taken to reach $\mathrm{AB}$ will be minimum.