Given equation of line are,
$l_{1}: x-3 y=0 $,Slope $m_{1}=\frac{1}{3}$
$l _{2}: 4 x +3 y =5$ slope $m _{2}=-\frac{4}{3}$
$l _{3}: 3 x + y =0$ slope $m _{3}=-3$
Here $m _{1} \times m _{3}=-1$ Line $l _{1} \perp l _{3}$
Slope of given line: $3 x -4 y =0, m =\frac{3}{4}$
herem $\times m_{2}=-1$
$\therefore $ Given line is perpendicular to $\text{Linel} _{2}$
Therefore The given line must pass throigh the orthocenter