Cyclic System of Equations

Multiply the first equation by $bc$, the second by $ca$, and the third by $ab$:
$$
\begin{array}{ll}
abc+1 = \dfrac{7}{6}bc\\
abc+1 = \dfrac{7}{3}ac\\
abc+1 = \dfrac{7}{2}ab
\end{array}
$$
Since each expression is equal to $abc+1$, we can set up another equation:
$$bc=2ac=3ab$$
This means that the ratio $a:b:c = 1:2:3$. We can now substitute $b$ and $c$ and write it in terms of a: $b=2a, c=3a$. Plugging it back into the system, we get
$$
\begin{array}{ll}
a+\dfrac{1}{6a^2} = \dfrac{7}{6}\\
2a+\dfrac{1}{3a^2} = \dfrac{7}{3}\\
3a+\dfrac{1}{2a^2} = \dfrac{7}{2}
\end{array}
$$
These are the same equation, so the final equation for $a$ is $$6a+\dfrac{1}{a^2} = 7$$, or $$6a^3-7a^2+1=0$$
We can factor the cubic into
$$
(a-1)(2a-1)(3a+1)=0
$$
Therefore, $$a=1, \dfrac{1}{2}, -\dfrac{1}{3}$$ As $abc=6a^3$, $$abc=\boxed{6,\dfrac{3}{4},-\dfrac{2}{9}}$$