Just to put things into the schematic editor here, find the following as Figure 2.10B:
simulate this circuit– Schematic created using CircuitLab
"because only \$\approx 0.6\: \text{mA}\$ of the \$4.4\:\text{mA}\$ collector current comes from \$R_3\$ -- make sure you understand why"
When looking at a circuit like this, your eye should be drawn to \$Q_3\$ and recognize it as a switch-mode use and not an analog-mode. This means \$Q_3\$ will be operated saturated, forcing the collector to act like a voltage source and to be brought very close to its emitter voltage, which is \$+15\:\text{V}\$. So expect the collector to be about \$+14.8\:\text{V}\$, as seen by the load.
Also, \$\beta_3\$ will be very low in such cases. Likely, \$\beta_3\lt 20\$. And perhaps for heavier loads perhaps as bad as \$\beta\lt 5\$.
This would be the case, perhaps, for a 2N3055:
So the next thought you have should be, "What's the load current and what kind of transistor must be used for \$Q_3\$?" The reason is that for larger currents the bipolar will have lower values for \$\beta\$, such as \$\beta=5\$, than would be the case for smaller currents, where these different bipolars could reasonably be expected to exhibit \$\beta=10\$ and \$\beta=20\$, when saturated.
Your eye should now be drawn to \$Q_2\$, which is also being used as a switch-mode device. Again, in saturation. Again, the same ideas about the collector behavior and its proximity to its emitter (which is at ground.) But in this case, it's much less likely that \$Q_2\$ will be carrying a high load current -- just the base current for \$Q_3\$, instead -- so this means that the expected \$\beta_2\$ can be closer to \$\beta_2=10\$ and \$\beta_2=20\$, when saturated.
At this point, you should have realized that the base of \$Q_3\$, when activated as a switch, will be somewhere between \$+14.4\:\text{V}\$ and \$+14.1\:\text{V}\$. (Where exactly will depend on details of \$Q_3\$ and how it is operated by this circuit.) And also, that \$Q_2\$'s collector will be close to ground. Perhaps \$\le +200\:\text{mV}\$. This means that there will be a difference between the base of \$Q_3\$ and the collector of \$Q_2\$ of about \$14\:\text{V}\$. This is something else that should be now entering your mind, when reading this circuit.
So, if you know the load's requirements as \$I_{_\text{LOAD}}\$, then it follows that \$R_2=\beta_3\,\cdot\frac{14\:\text{V}}{I_{_\text{LOAD}}}\$. Their value of \$R_2=3.3\:\text{k}\Omega\$ means a \$Q_2\$ collector current of about \$\frac{14\:\text{V}}{3.3\:\text{k}\Omega}\approx 4.3\:\text{mA}\$. This is about what the textbook says.
This is low enough that we can say that \$Q_3\$ is not a 2N3055.
At this point it is time to reflect.
This circuit will support \$I_{_\text{LOAD}}\le 100\:\text{mA}\$. This means that \$Q_3\$ is smaller, not larger. So \$\beta_3\ge 20\$. Likely, anyway.
From the above, we can also expect that \$\beta_2\ge 20\$, as well. This means that that the base current for \$Q_2\$ will be \$\le 250\:\mu\text{A}\$. From this, we can work out that \$R_1\approx \frac{3.3\:\text{V}-650\:\text{mV}}{250\:\mu\text{A}}\approx 10\:\text{k}\Omega\$. Take note that this matches their selection quite closely!
This leaves \$R_3\$. Here, the "rule of thumb"(my apologies to women and the historical origins of that phrase) would be to use about 10% of the \$4.3\:\text{mA}\$ for this pull-up. So we'd guess at \$R_3\approx\frac{800\:\text{mV}}{430\:\mu\text{A}}=1.86\:\text{k}\Omega\$. Making this a little stiffer would mean we could accept \$R_3\approx 1\:\text{k}\Omega\$.
At this point, we've decided to agree with the authors. And we know a lot more about the range of the allowable load, now.
What happens if \$R_3=100\:\Omega\$?? Well, the current in \$R_3\$ will increase by about a factor of 10. Instead of about \$430\:\mu\text{A}\$, it will move towards about \$4.3\:\text{mA}\$. Which is all of our allocated base current for \$Q_3\$.
So that is why lowering \$R_3=100\:\Omega\$ is a problem. It would siphon away the necessary base current for \$Q_3\$ when that low.
Hopefully, this mental walk through the circuit helps.