According to what I learnt the direction ratio $(3, -2, 8)$ is same as that of the direction ratio $(-3, 2, -8)$.
Again we know that, if $l$, $m$, $n$ are direction cosines and $a$, $b$, $c$ are direction ratios then,
$l =\displaystyle \frac{a}{\sqrt{a^2 + b^2 + c^2}}$, $m =\displaystyle \frac{b}{\sqrt{a^2 + b^2 + c^2}}$ and $n =\displaystyle \frac{c}{\sqrt{a^2 + b^2 + c^2}}$
So, the two direction ratios will give two different direction cosines.Again I learnt that the direction cosines of a line is unique. So surely the two different direction cosines will represent two different lines.
I think here both the lines represented by the two direction cosines are the same, only their their direction is opposite to that of each other. Am I correct?
The direction cosines are defined for a vector $V \in \mathbb{R}^3 $ not a line.
If $ V = [v_1, v_2, v_3]$ , and if $ a, b, c $ are the angles that the vector $V$ makes with the $x, y, z$ axes, then the direction cosines are given by
$ \alpha = \cos(a) = \dfrac{ v_1 }{ \sqrt{v_1^2 + v_2^2 + v_3^2}} $
$ \beta = \cos(b) = \dfrac{ v_2 }{ \sqrt{v_1^2 + v_2^2 + v_3^2} }$
$ \gamma = \cos(c) = \dfrac{ v_3 }{ \sqrt{v_1^2 + v_2^2 + v_3^2}} $
From these equations, you can see that if you reverse $V$ into another vector $U = -V$, then the direction cosines of $U$ will be the negative of those of $V$, with corresponding changes to the angles $a,b,c$.
The direction ratios of a vector are just the coordinates of this vector. So if reverse the vector $V$ into $U = -V$ , the direction ratios will be the negative of those of $V$.
So the answer to your question in the title of this question is YES, taking the negative of the direction ratios (of a vector) changes the direction cosines of the resulting vector to the negative of the original direction cosines.