Suppose you have a spin-1 system. Let us resonantly drive the transition between any 2 levels (say 0 1 transition). How would the the presence of the third level (-1) state affect this transition?
We know that there will be a certain amount of off resonant transition between 0 and -1 states. So can we write a spin hamiltonian that essentially is an interaction between the 0 1 transition and 0 -1 levels?
There are several ways to approach this answer. First of all, if you consider a spin-1 system isolated in free space, the three $M_S$ sublevels will be degenerate. So I assume you imagine placing the spin in an external magnetic field. In this case, for a true spin-1 system the Zeeman shifts will be like $E_{M_S} \propto M_S$, i.e. the splitting between $M=0$ and $M=1$ will exactly match that between $M=0$ and $M=-1$. That is to say these two transitions will share the same resonance frequency. If you resonantly drive one of these transitions, you will simultaneously drive the second transition. This is not because of off-resonant excitation of the transition you didn't intend to drive; both transitions have exactly the same resonance frequency, after all! So, you should represent the system by a full $3\times3$ matrix including both the energy levels in the absence of the driving field and the couplings due to the driving field and solve the full system without approximating it as two disjoint parts.
It has no effects if your applied driving has no matrix element to the third level (most likely due to symmetry). Otherwise, under RWA, you can still write a $3\times3$ matrix as the other answer has pointed out. Based on your description, the Hamiltonian is likely to be $\left(\begin{matrix}0 & \Omega_1 & 0 \\ \Omega_1 & \delta & \Omega_2\\0 & \Omega_2 & \Delta\end{matrix}\right)$, where $\delta\sim\Omega_1$ is the detuning you used to have for the 2-level transition, and since you said "resonantly driving" it means $\delta=0$, but generally I would like to keep it here. $\Delta$ is the detuning from the transition to the other level. If $\Delta\gg\Omega_2$, then this would only act as an AC stark shift on the middle transition by $-\Omega_2^2/\Delta$, with a higher-order correction to $\Omega_1$ due to the mixing: $H_\mathrm{eff}=\left(\begin{matrix}0 & \Omega_1\\ \Omega_1 & \delta -\Omega_2^2/\Delta\end{matrix}\right)$.
