Explanation

(b)
The point of intersection of lines $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$ is
$\left( - \frac{2}{34},\frac{22}{17} \right)$
The slope of required line which is perpendicular to
$6x - 7y + 3 = 0\ is - \frac{7}{6}$
$\therefore$ Equation of required line
$y - \frac{22}{17} = - \frac{7}{6}\left( x + \frac{2}{34} \right)$
$\Rightarrow \frac{6(17y - 22)}{17} = - \frac{7(34x + 2)}{34}$
$\Rightarrow 119x + 102y = 125$