If two straight lines whose direction cosines are given by relation $l+m-n=0,3l^2+m^2+cnl=0$ are parallel then the positive value of  ‘c’

$l+m-n=0\to \left(1\right)3l^2+m^2+Cnl=0\to \left(2\right)\text{Sub (1) in (2)}$

$3l^2+m^2+cl(l+m)=0(3+c){\left(\frac{l}{m}\right)}^2+c\left(\frac{l}{m}\right)+1=0$

$\because$lines  are parallel  So roots must be equal $\Rightarrow D=0c^2-4(3+1)=0c=6 \text{or }c=-2$