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I've encountered the following circuit and I'm puzzled about the effective resistance that the opamp will see.
I know that R1 is in parallel with R2+R3, so this would be easily answered by R1*(R2+R3)/R1+(R2+R3). But there's a ground between R2 and R3 and I don't know if that will affect the total resistance the opamp will see.
I know that R1 is in parallel with R2+R3
No it isn't; R2 and R3 both connect to ground.
I'm puzzled about the effective resistance that the opamp will see.
I'm assuming you mean the resistance/impedance present at the inverting input node.
Because it's a virtual ground (due to negative feedback), the op-amp's inverting input (if it had feelings or the ability to understand impedance) would "detect" an impedance of near-zero-ohms.
It's a virtual ground because it's an integrator with negative feedback from output to inverting input and, we can assume that the op-amp has a massive open-loop gain hence, the voltage at the inverting input will, under normal operational conditions, be at the same voltage as the non-inverting input (ground potential).
If in fact you meant "what's the effective input resistor for the integrator" then, the answer is 68 kΩ. R3 is nulled-out by the virtual ground and, for this circuit to work as intended, there will be a voltage source on the left hand node pointing off the page. This voltage source will have a very low output impedance (ideally zero) and therefore, any effect that R2 has is lost hence, R1 is the effective impedance between the unseen voltage source (left of page) and the inverting input.
