$4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c$
$\equiv(2 a+ b)^{2}-3(2 a+ b) c+2 c^{2}=0$
$\Rightarrow (2 a+b-2 c)(2 a+ b -c)=0$
$\Rightarrow c=2 a+b$ or $c=a+\frac{1}{2} b$
The equation of the family of lines is
$a(x+2)+b(y+1)=0$
or $a(x+1)+b\left(y+\frac{1}{2}\right)=0$
giving the point of concurrence $(-2,-1)$
or $\left(-1,-\frac{1}{2}\right)$
$a(x+2)+b(y+1)=0$
or $a(x+1) +b\left(y+\frac{1}{2}\right)=0$
giving the point of concurrence
$(-2,-1)$ or $\left(-1,-\frac{1}{2}\right)$