Solution:
We know $p q r = 48$ since the product of the roots of a cubic is the constant term. Now,
$$\frac{1}{p} + \frac{2}{q} + \frac{3}{r} \geq 3 \sqrt[3]{\frac{6}{p q r}} = \frac{3}{2}$$
by AM-GM, with equality when $1/p = 2/q = 3/r$. This occurs when $p = 2$, $q = 4$, $r = 6$, so $3/2$ is in fact the minimum possible value.