Let $$a,b,c$$ be reals satisfying $$3ab+2=6b$$, $$3bc+2=5c$$, and $$3ca+2=4a$$. Let $$\mathbb{Q}$$ denote the set of all rational numbers. Given that the product $$abc$$ can take two values $$\frac{r}{s}\in\mathbb{Q}$$ and $$\frac{t}{u}\in\mathbb{Q}$$, in lowest form, find $$r+s+t+u$$.

We are given the three simultaneous equations

$$3ab+2=6b,\qquad 3bc+2=5c,\qquad 3ca+2=4a$$

Move the constant term to the right in each equation and factor out the common variable:

$$b(3a-6)=-2 \quad -(1)$$
$$c(3b-5)=-2 \quad -(2)$$
$$a(3c-4)=-2 \quad -(3)$$

From $$-(1)$$, express $$b$$ in terms of $$a$$:

$$b=\frac{-2}{3a-6}\qquad\text{provided }3a-6\neq0$$

Insert this $$b$$ into $$-(2)$$ to get $$c$$ in terms of $$a$$:

$$c=\frac{-2}{3b-5}=\frac{-2}{3\left(\frac{-2}{3a-6}\right)-5} =\frac{-2(3a-6)}{24-15a}\qquad -(4)$$

Now place this $$c$$ into $$-(3)$$ to obtain an equation involving only $$a$$:

$$a=\frac{-2}{3c-4} =\frac{-2}{\frac{-6(3a-6)}{24-15a}-4}$$

Combine the terms in the denominator:

$$3c-4=\frac{-6(3a-6)-4(24-15a)}{24-15a} =\frac{42a-60}{24-15a}$$

Therefore

$$a=\frac{-2(24-15a)}{42a-60}$$

Cross-multiplying gives the quadratic in $$a$$:

$$42a^{2}-60a=-48+30a$$
$$42a^{2}-90a+48=0$$
$$7a^{2}-15a+8=0$$

The discriminant is $$\Delta =(-15)^{2}-4\cdot7\cdot8=225-224=1$$, so

$$a=\frac{15\pm1}{14}\;\Longrightarrow\; a=\frac{8}{7}\quad\text{or}\quad a=1$$

If $$a=\dfrac{8}{7}$$,
$$b=\frac{-2}{3a-6}=\frac{-2}{\frac{24}{7}-6}=\frac{7}{9},\quad c=\frac{-2}{3b-5}=\frac{-2}{\frac{7}{3}-5}=\frac{3}{4}$$

Product: $$abc=\frac{8}{7}\cdot\frac{7}{9}\cdot\frac{3}{4}=\frac{2}{3}$$

If $$a=1$$,
$$b=\frac{-2}{3-6}=\frac{2}{3},\quad c=\frac{-2}{3b-5}=\frac{-2}{2-5}=\frac{2}{3}$$

Product: $$abc=1\cdot\frac{2}{3}\cdot\frac{2}{3}=\frac{4}{9}$$

The two attainable values of $$abc$$ are $$\dfrac{2}{3}$$ and $$\dfrac{4}{9}$$. Writing them in lowest terms as $$\dfrac{r}{s}$$ and $$\dfrac{t}{u}$$ we have

$$r=2,\; s=3,\; t=4,\; u=9$$

Hence $$r+s+t+u = 2+3+4+9 = 18$$

Final Answer: 18