If one of the lines given by $6x^2-xy+4c{y}^2=0$ is $3x+4y=0,$ then $c$ equals 

It is given that $3x+4y=0$ i.e. $y=-\frac{3}{4}x$

the lines given by $6x^2-xy+4c{y}^2=0.$ There, $y=-\frac{3}{4}x$

satisfies it.

$\therefore 6x^2+\frac{3}{4}{x}^2+4c\times \frac{9}{16}{x}^2=0\Rightarrow \frac{27}{4}+\frac{9c}{4}=0\Rightarrow c=-3$