Well,
LOAD A and R2 are submitted to the same potential (12V), because they are in parallel.
LOAD B and R3 follow the same case above, but they are submitted to - 20 V.
Current of LOAD A + R2 = Current of LOAD B + R3 = Current of R1 (they are in parallel).
The total battery voltage is 35 V. So, if tou hava a 12V drop at LOAD A, a 20V drop at LOAD B, your drop at R1 will be 35 - (12 + 20) = 3V
Following the KCL and KVL:
12/R2 + 23.53mA = 20/R3 + 17.65 mA
12/R2 - 20/R3 = -5.88 mA (If we substract the current passing through R3 from the current passing through R2 we must have a 5.88 mA excess, to balance the currents).
Now, if I understood the bleeder rule:
(23.53 + 17.65)*0.1 = 4.118 mA
Let's take that value as the reference, as far I understand, you need at least a 10% of current bleeding, so:
IR2 = 4.118 mA
IR3 = 4.118 mA + 5.88 mA = 9.998 mA
Total current at LOAD A and R2 = 4.118 + 23.53 = 27.648 mA
Total current at LOAD B and R3 = 9.998 + 17.65 = 27.648 mA
KCL met!
R2 value = 12V/4.118mA =~ 2,914 ohms
R3 value = 20V/9.998mA =~ 2,000.4 ohms
R1 value = 3V/27.648mA =~ 108.5 ohms