$3x+4y=0$
is one of the lines of the pair.
$6x^2-xy+4cy^2=0$
Put $y=-\large\frac{-3}{4}$$x$
We get $6x^2+\large\frac{3}{4}$$x^2+4c\bigg(\large\frac{-3}{4}$$x^2 \bigg)=0$
=> $6+\large\frac{3}{4} +\frac{9c}{4}$$=0$
$c=-3$
Hence A is the correct answer.