If two straight lines whose direction cosines are given by the relations $\mathcal{l}+m-n=0, 3{\mathcal{l}}^2+m^2+c\mathcal{l}n=0$ are parallel, then the sum of possible values of c is

$\mathcal{l}+m-n=0\Rightarrow \mathcal{l}+m=n$ 
$3{\mathcal{l}}^2+m^2+c\mathcal{l}n=0\Rightarrow 3{\mathcal{l}}^2+m^2+c\mathcal{l}(\mathcal{l}+m)=0$ 
$\Rightarrow (3+c){\mathcal{l}}^2+m^2+c\mathcal{l}m=0$ 
$\Rightarrow (3+c)\frac{{\mathcal{l}}^2}{m^2}+c\frac{\mathcal{l}}{m}+1=0$  ……… (1)
Lines are parallel   root of (1) are equal
$\Rightarrow \Delta =0$ 
$\Rightarrow c^2-4(3+c)=0\Rightarrow c^2-4c-12=0$ 
$\Rightarrow c=6,-2$ 
Sum of possible values of c is $6+(-2)=4$