Since theoretically solving it will be a complex task...
I agree but such problems are not at all uncommon. Many of us will go for a simulation but I love the good old fashioned graphical solution. It is simple and gives a very open view into the operation.
I can draw but I am lazy! So I shall just outline the steps:
1. R1 and R2 are in series; they can be combined. Same is true for R3 and R4.
2. If you look hard, you can see that the same current must always flow via these two pairs: hence they can be combined further: we replace R1, R2, R3 and R4 with one resistor 712K...
3. The input 220V AC (assume RMS: peak 310V) is rectified by the bridge of four diodes. Ignore diode drops and plot the result. You will get the following graph:
View attachment case1.pdf
4. Now add the R5 as load; the curve will not change but the peak will come down by some: 310*102/(102+712)=39V this will be the max voltage once we put the 102K load. The graph will remain same in form but the peak will become 39V.
5. Now put the capacitor; the capacitor will charge max to 39V (actually less) but the Xc for the cap has to be calculated at 100Hz. You will be charging the cap at const curr approx 310/712mA=0.4mA
6. The sine curve will be distorted (look more like a ramp) and it goes upto 39V.
7. Next branch is left as an exercise for the student. You have to divide the voltage 0V to 30V and 30V to up (two steps)