If a line (x+1)/(p) = (y-1)/(q) = (z-2)/(r) where p = 2q = 3r ,...

The direction ratios of the line are proportional to $p, q, r$.

Given $p = 2q = 3r$.

Let $p = 2q = 3r = 6k$.

Then $p = 6k$, $q = 3k$, $r = 2k$.

The direction ratios of the line are proportional to $6, 3, 2$.

The direction cosines of the line are $\dfrac{6}{\sqrt{6^2+3^2+2^2}}, \dfrac{3}{\sqrt{6^2+3^2+2^2}}, \dfrac{2}{\sqrt{6^2+3^2+2^2}}$, which simplifies to $\dfrac{6}{7}, \dfrac{3}{7}, \dfrac{2}{7}$.

The line makes an angle $\theta$ with the positive direction of the $y$-axis, so $\cos \theta = \dfrac{3}{7}$.

Using the double angle formula, $\cos 2\theta = 2\cos^2 \theta - 1$.

$\cos 2\theta = 2\left(\dfrac{3}{7}\right)^2 - 1 = \dfrac{18}{49} - 1 = -\dfrac{31}{49}$.

Answer: $-\dfrac{31}{49}$