GIVEN:

$2x-3y+4=0$…(i)

$3x+4y-5=0$…(ii)

$6x-7y+8=0$…(iii)

Here the individual is remaining at the intersection of the ways addressed by lines (1) and (2).

By settling conditions (1) and (2) we get $x=1/7$ and $y=22/17$

Subsequently, the individual is remaining at point $(-1/17,22/17)$

We realize that the individual can arrive at way (3) at all time on the off chance that he strolls along the opposite line to (3) from point $(-1/17,22/17)$

Here the slant of the line (3) $=6/7$

We get the slant of the line opposite to line (3) $=-1/(6/7)=7/6$

So the condition of line going through $(-1/17,22/17)$ and having a slant of $-7/6$ is composed as

By additional estimation $\left( y-\frac{22}{17} \right)=-\frac{7}{6}\left( x+\frac{1}{17} \right)$

$6(17y+22)=7(17x+1)$

By increase $102y-132=119x+7$

We get $119x+102y=125$

Accordingly, the way that the individual ought to follow is $119x+102y=125$.