I've only done this for 2D systems, never for 3D.
Find the equilibrium points for this system of equations (your answer will depend on the value of $N=S+I+R$, which you may assume to be a constant). ($a,b,c$ are positive constants.) $$\dot S=cR - aSI$$ $$\dot I = aSI - bI$$ $$\dot R = bI - cR$$
My attempt:
From $\dot I = aSI - bI$, if $\dot I =0$, then $aSI = bI \therefore I = 0$ or $S = \frac{b}{a}$
If $I = 0$, then by $\dot R$, $R=0$, and $S$ could be any value, therefore $(S,I,R)=(k,0,0)$ is a fixed point for any $k \in [0,N]$.
Is this remotely somwhere along the lines of what this question is asking?
Yes, you're following exactly the right approach. Now you can try to find any other equilibrium points by taking the second choice which yields $\dot{I} = 0$, namely $S = \frac{b}{a}$, and repeat the line of reasoning you already employed. Hint: you will obtain two distinct 'lines' of equilibria (you already found one of them).