The middile one definitely has something wrong. You don't use a resistor potential divider as a power supply [unless you use a power op-amp to prevent (make negligible) the effects of loading, but that is a whole other kettle of fish].
You should use a proper voltage regulator - be it a LDO regulator, a switching regulator, a zener diode, or other such voltage regulation devices
Loading 101:
If say for example you have two 1k resistors as a divider and say connect 5v at the top. Under no load, you will get 2.5V. Say you now used that as a power supply, and started drawing i dont know, maybe 1mA from the 2.5V line
You would then have
Ib = Vb/Rb = Vb/1000 [A] flowing through the bottom resistor, and 1mA flowing through the output.
That would then mean that flowing through the top resistor you have:
I = 1 + Vb [mA] = 0.001 + 0.001Vb [A]
The voltage drop on the top resistor would then be:
Vt = IR = 1000*I = 1 + Vb [V]
That would give you a total voltage drop across the potential divider:
V = 5[V] = Vt + Vb = 1+2Vb
So that would mean that at the output, the voltage would be:
Vb = (5 - 1)/2 = 4/2 = 2[V]
So in that case, just 1mA of loading would result in the output voltage dropping by 20%.
In your case you seem to be using 10k for the lower resistor and 5k for the upper resistor and are attempting to power 3 LEDs from it. So input voltage would be 5V, load current would be say 60mA. So just for fun that would give Vb = -200[V]. That figure is of course absurd, and the reason for it is you simply cannot put 60mA through a 5k resistor when the voltage across it is limited to 5V (The maximum would be I=V/R = 5/5000 = 1[mA], and at that amount there would be 0v across the LED).