Explanation

(c)
The point of intersection of lines $x - 2y$ and $x + 3y = 2$ is
$\left( \frac{7}{5},\frac{1}{5} \right)$
Since, required line is parallel to $3x + 4y = 0$.
Therefore, the slope of required line $= - \frac{3}{4}$
$\therefore\$ Equation of required line whose slope is $\frac{- 3}{4}$ and passes through $(\frac{7}{5},\frac{1}{5})$ is
$y - \frac{1}{5} = - \frac{3}{4}\left( x - \frac{7}{5} \right)$
$\Rightarrow 20y - 4 = - 15x + 21\ \Rightarrow 3x + 4y - 5 = 0$