The intersection point of lines $x - 2y = 1$ 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 is $ - \frac{3}{4}$
$\therefore $ Equation of required line which passes throught $\left( \frac{7}{5}, \frac{1}{5}\right)$ is given by
$y - \frac{1}{5} = - \frac{3}{4}\left(x - \frac{7}{5}\right)$
$ \Rightarrow \frac{3x}{4} + y = \frac{21}{20} + \frac{1}{5}$
$\Rightarrow 3x+ 4y - 5 = 0 $