20. Let $a, b, c$ be real numbers such that
$$
3 a b+2=6 b, \quad 3 b c+2=5 c, \quad 3 c a+2=4 a .
$$

Suppose the only possible values for the product $a b c$ are $r / s$ and $t / u$, where $r / s$ and $t / u$ are both fractions in lowest terms. Find $r+s+t+u$.
20. 18
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 $r+s+t+u=4+9+2+3=18$.