Thanks a lot. I managed to get the expected right answer for this and your values are much closer to it:

R2 = 68 Ω
R3 = 100 Ω
R6 = 33 Ω .

I'm reviewing my calculations now.
Thanks a lot for all the help.

I have no idea how there can be any "expected right answers". I picked a value of R3 arbitrarily --- I happened to choose a value that made that branch have a resistance of 180 Ω so that the parallel equivalent of that branch and R5 was trivial to arrive at.

There is a reason I asked, way back in Post #2:

When the switch is open, what constraints can you come up with for the values of R2 and R3?

You simply can't come up with a single set of "correct" values. But you CAN come up with bounds on what each value needs to be and also on the relationship that must exist between the two of them.

Before we start doing this, all we know is that

0 ≤ R2 ≤ ∞
0 ≤ R3 ≤ ∞

Our goal is to narrow this down as aggressively as possible.

If R2 is too large, then Vab will be driven to a small value regardless of what R3 is. The limiting value will be when R3 = ∞, which will require that R2 ≤ 155.62 Ω.

Notice that the choice you made is relatively close to this value and that's because your choice for R2 is, relative to the other resistances in this circuit, essentially infinite.

On the other hand, if R3 = 0 Ω, then R2 = 36.77 Ω. That means that

36.77 Ω ≤ R2 ≤ 155.62 Ω

Note that I'm carrying extra sig figs because these can be viewed as potentially intermediate results.

You can now analyze the circuit to determine the relationship between R2 and R3 which will let you put limits on R3 (in light of the limits on R2).

You can then do the same thing for the case when the switch is closed, but its more involved, to find relationships between two of the resistances in terms of the third as well as the limits on all three.

You are then free to pick any set of values that satisfy those constraints.

Not all of those choices are necessarily equal as the sensitivity of the voltage Vab (or whatever other parameter is of particular interest) to changes/uncertainties in the three resistances is a function of that choice -- but that is WAY beyond what you are being asked for in this problem. Just something to think of as a teaser for later study.

EDIT: Errors pointed out by The Electrician corrected.

 

If R2 is too large, then Vab will be driven to a small value regardless of what R3 is. The limiting value will be when R3 = ∞, which will require that R2 ≤ 177.62 Ω.

I get a different result.

If R3= ∞ and S1 is open, then all we have left is R1, R2 and R5 across the 12 volt supply. R1+R2 and R5 form a voltage divider, and if we want 6.04 volts across R5 we need 180/(R2+22+180) = 6.04/12
which gives R2 = 155.616Ω

I also get a different result when R3 = 0.

 

The Vab = 6.04 V is for the S1 switch at open state.
When it is closed, then Vab = 4.03 V .

and

I managed to get the expected right answer for this and your values are much closer to it:

R2 = 68 Ω
R3 = 100 Ω
R6 = 33 Ω .

you can't reconcile those two sets of numbers

 

I get a different result.

If R3= ∞ and S1 is open, then all we have left is R1, R2 and R5 across the 12 volt supply. R1+R2 and R5 form a voltage divider, and if we want 6.04 volts across R5 we need 180/(R2+22+180) = 6.04/12
which gives R2 = 155.616Ω

I also get a different result when R3 = 0.

That's what I get for doing everything directly on the calculator and not writing any work down as I went -- didn't have paper/pen handy.

At first glance, it appears I forgot to subtract off the 22 Ω for the upper limit. Since the voltage is just barely over half the supply voltage, we need the combination of R2 and 22 Ω to be just barely under 180 Ω, so that jives with my 177.62 Ω result. Subtracting the 22 Ω yields 155.62 Ω, which matches your result.

I possibly did the same thing for the other limit. If R3 = 0 Ω, then the resistance between a and b is 180 Ω || 89 Ω = 59.55 Ω. Which I'm thinking is the value I put in the memory and then failed to overwrite it with the final result. This makes the required value of R2 (after taking the 22 Ω into account) 36.77 Ω (which sounds familiar, so I think I just got confused regarding what value was actually in the memory).

Thanks for pointing out the errors.