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

Question

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

Accepted Answer

With $n=l+m$: $3l^2+m^2+cl(l+m)=0 \implies (3+c)l^2+clm+m^2=0$.

Treating as a quadratic in $l/m$, the two lines are parallel exactly when this has a repeated root:
$$c^2-4(3+c)=0 \implies c^2-4c-12=0 \implies (c-6)(c+2)=0.$$
Positive value: $c=6$.

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