If two straight lines whose direction cosines are given by the relations l + m - n = 0 , 3 l 2 + m 2 + c n l…

Given, $l+m-n=0$ & $3l^2+m^2+cln=0$

Now taking $n=l+m$ and putting in $3l^2+m^2+cln=0$ we get,

$3l^2+m^2+cl\left(l+m\right)=0$

$\Rightarrow 3l^2+m^2+c{l}^2+clm=0$

$\Rightarrow \left(3+c\right)l^2+clm+m^2=0$

$\Rightarrow \left(3+c\right){\left(\frac{l}{m}\right)}^2+c\left(\frac{l}{m}\right)+1=0$ ..equation (i)

$\because$  lies are parallel.

So, roots of equation (i) must be equal

$\Rightarrow D=0$

$\Rightarrow c^2-4\left(3+c\right)=0$

$\Rightarrow c^2-4c-12=0$

$\Rightarrow \left(c-6\right)\left(c+2\right)=0$

$c=6$ or $c=-2$

So, positive value of $c=6$