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

The direction cosines satisfy: $$l + m - n = 0$$ and $$3l^2 + m^2 + cnl = 0$$.

Since $$l + m - n = 0$$, we get $$n = l + m$$.

Substituting this into the second equation gives $$3l^2 + m^2 + c \cdot (l + m) \cdot l = 0$$, which simplifies to $$3l^2 + m^2 + cl^2 + clm = 0$$ and hence $$(3 + c)l^2 + clm + m^2 = 0$$.

Now dividing by $$m^2$$ and setting $$t = l/m$$ leads to $$(3 + c)t^2 + ct + 1 = 0$$.

For the two lines to be parallel, both direction cosine ratios must coincide, so the quadratic in $$t$$ must have equal roots, i.e.\ its discriminant vanishes. In other words, $$D = c^2 - 4(3 + c)(1) = 0$$, which simplifies to $$c^2 - 12 - 4c = 0$$ or $$c^2 - 4c - 12 = 0$$. Factorizing gives $$(c - 6)(c + 2) = 0$$, hence $$c = 6$$ or $$c = -2$$.

The positive value of $$c$$ is $$6$$.

The answer is Option A: $$6$$.