Thank you for reporting, we will resolve it shortly
Q.
If two straight lines whose direction cosines are given by the relations $1+ m - n =0,3l^{2}+ m ^{2}+ cnl =0$ are parallel, then the positive value of $c$ is :
$1+ m - n =0$
$3 l ^{2}+ m ^{2}+ cl (1+ m )=0$
$n =1+ m$
$31^{2}+ m ^{2}+ cl ^{2}+ clm =0$
$(3+ c ) l ^{2}+ clm + m ^{2}=0$
$(3+c)\left(\frac{l}{m}\right)^{2}+c\left(\frac{l}{m}\right)+1=0 \ldots \ldots(1)$
$\because$ lies are parallel.
Roots of (1) must be equal
$\Rightarrow D=0$
$c^{2}-4(3+c)=0$
$c^{2}-4 c-12=0$
$(c-6)(c+2)=0$
$c=6 \text { or } c=-2$
+ve value of $c=6$