Equilibrium of more than two inter reacting species
At equilibrium $~r_1~=r_2~=r_3~=~0$.
So, $$k_1[A]-k_{-1}[B]=0$$
$$k_2[B]-k_{-2}[C]=0$$
$$k_3[C]-k_{-3}[A]=0$$
So, The criteria for equilibrium is
1) $$k_1[A]=k_{-1}[B]$$
2) $$k_2[B]=k_{-2}[C]$$
3) $$k_3[C]=k_{-3}[A]$$
Now defining equilibrium constant as $K_i=\frac{k_i}{k_{-i}}$
We can derive a formula
$$K_1 \times K_2 \times K_3~=~1$$