Explanation

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