Solution:
The three given equations can be written as

$$3 a + \frac{2}{b} = 12, \quad 3 b + \frac{2}{c} = 10, \quad 3 c + \frac{2}{a} = 8$$

The product of all the three equations gives us
$$27 a b c + 6\left(3 a + \frac{2}{b}\right) + 6\left(3 b + \frac{2}{c}\right) + 6\left(3 c + \frac{2}{a}\right) + \frac{8}{a b c} = 120$$

Plugging in the values and simplifying the equation gives us
$$27(a b c)^2 - 30 a b c + 8 = 0$$
This gives $a b c$ as either $4 / 9$ or $2 / 3$, so $r+s+t+u=4+9+2+3=18$.