Let the direction cosines of two lines satisfy the equations : 4l + m - n =0 and 2mn + l0nl +3lm= 0. Then the cosine of the acute angle between these lines is :

We are given two equations for direction cosines $$(l, m, n)$$: $$4l + m - n = 0\quad\cdots(1)$$ and $$2mn + 10nl + 3lm = 0\quad\cdots(2)$$. (Note: “l0nl” in the problem is interpreted as “10nl”.)

First, from (1) we have $$n = 4l + m$$, and substituting this into (2) gives $$2m(4l + m) + 10(4l + m)l + 3lm = 0$$, which simplifies to $$8lm + 2m^2 + 40l^2 + 10lm + 3lm = 0$$ and hence $$40l^2 + 21lm + 2m^2 = 0$$.

Next, dividing by $$m^2$$ and setting $$t = l/m$$ yields the quadratic equation $$40t^2 + 21t + 2 = 0$$. Solving gives $$t = \frac{-21 \pm \sqrt{441 - 320}}{80} = \frac{-21 \pm \sqrt{121}}{80} = \frac{-21 \pm 11}{80}$$, so $$t_1 = -\frac{1}{8}$$ and $$t_2 = -\frac{2}{5}$$.

For the first line, where $$l/m = -\frac{1}{8}$$, we may choose $$l = -1$$ and $$m = 8$$, which leads to $$n = 4(-1) + 8 = 4$$, giving the direction ratios $$(-1, 8, 4)$$.

Similarly, for the second line with $$l/m = -\frac{2}{5}$$, setting $$l = -2$$ and $$m = 5$$ yields $$n = 4(-2) + 5 = -3$$, so the direction ratios are $$(-2, 5, -3)$$.

Finally, the cosine of the acute angle between these two vectors is $$\cos\theta = \frac{|(-1)(-2) + (8)(5) + (4)(-3)|}{\sqrt{1 + 64 + 16}\,\sqrt{4 + 25 + 9}} = \frac{|2 + 40 - 12|}{\sqrt{81}\,\sqrt{38}} = \frac{30}{9\sqrt{38}} = \frac{10}{3\sqrt{38}}$$.

The correct answer is Option 3: $$\frac{10}{3\sqrt{38}}$$.