If  $a^2+9b^2-16c^2=6abandax+by+c=0$  represents a family of concurrent lines then its point of Intersection is 

$a^2+9b^2-6ab=16c^2$

$(a-3b{)}^2=(4c{)}^2$

$a-3b=\pm 4c$

$a-3b-4c=0$

$\frac{a-3b}{4}-c=0$

$\frac{-a}{4}+\frac{3}{4}b+c=0$

$compare with ax+by+c=0$

$(x,y)=(-\frac{1}{4},\frac{3}{4})$

and $a-3b+4c=0$

$\frac{a}{4}-\frac{3b}{4}+c=0$

$compare with ax+by+c=0$

$(x,y)=(\frac{1}{4},\frac{-3}{4})$