Working of the below Op-amp circuit
I'll not address the AC behavior of this circuit- I suspect the AC characteristics might not have been 'designed' as such anyway. So ignore the capacitors.
You can equate the voltages at the inputs of the op-amp.
The voltage at the non-inverting input is simply Vload (we can ignore offset voltage and input bias current- note that the designer has reasonably balanced the impedances seen by the two inputs, though 20K would be better for R4).
The voltage at the inverting input is more interesting- assuming the shunt regulator is regulating-
It's Vinv = \$V_{LOAD}+I_{LOAD} \cdot 2.34\Omega - 1.24V\cdot\frac{33K\Omega}{33K\Omega+47K\Omega}\$
So \$I_{LOAD} = k(1.24V/2.34\Omega)\$ where k = 33/(33+47) = 0.4125
or about 216mA.
The shunt regulator, as you can see from the above, is the reference for the constant current. By dividing it down by R3/(R1+R3) less voltage is dropped across the sense resistor R7.
R2 is there to provide adequate bias current for the shunt regulator, a job at which I can't see it quite doing at the lower bound for the load resistance.
216mA * (5.34 ohms) = 1.15V. The reference needs about 55 to 100uA to operate properly, so that's another 55-100mV, so it's unlikely the regulation will work properly below about 3.7 ohms. Upper limit for load resistance calculation is left as an exercise.