The Intuition for RLC circuits
Resistors are the electrical equivalent of friction and produce losses to remove energy from the system. If you drop a frictionless pendulum to swing it, how long does it take to settle? Infinite time, because there are no losses so it just swings back and forth and oscillates around bottom position forever. The flip side is what happens if the pendulum has infinite friction? Well it just sits there at the high position forever taking infinite time to reach the bottom forever hanging on to its potential energy.
If you have non-infinite but high friction, your pendelum is going to take a VERY long time to reach the bottom position. It will never overshoot but it will take a very long time to get there. And if you have too little it will reach the bottom point faster but overshoot and then proceed to swing continuously about the bottom position for a very long time too. There is a middle ground that mutually minimizes the time it takes for energy to dissipate and the time it takes for the pendelum to travel one cycle to the bottom position.
Resistor = friction. Dissipated as heat (for the most part) in both systems.
Inductor = kinetic energy storage. In the pendulum this is the velocity of the weight combined with the mass.
Capacitor = potential energy storage. In the pendulum this is the gravitation potential energy due to the mass being at a height.
You might be wondering why the capacitor is the potential energy storage and why the inductor is the kinetic energy storage. This is because you can put charge in the capacitor and then just keep the energy there statically by disconnecting it (like sealing air in a pressurized tank or just holding the pendulum at a height). But in an inductor it requires current constantly flowing through it to store energy in its magnetic field otherwise the field collapses (just like a pendulum requires constant motion to store kinetic energy). You can't store energy in an inductor or kinetic energy statically. You can only store it dynamically.
So in both a pendulum energy is being exchanged between kinetic and potential. The pendulum exchanges speed for height back and forth. In the RLC circuit, the capacitor is exchanging energy with the inductor. With overly low resistance (parallel RLC), high resistance (series RLC), or high friction it takes a very long time for energy to completely transfer from one medium to the other, but it overshoots very little once the transfer is complete because so much is dissipated during the transfer with only one transfer occurring at the extreme but taking a very long time to do so. With very high resistance (parallel RLC), low resistance (series RLC), or low friction, a single transfer completes very quickly but so little energy is lost that much of it remains to be transferred back in the other direction so many, many transfers can occur.
The corrected statement is that the slightly underdamped case has the shortest settling time.
Now why is this the case when it's obvious that a highly underdamped system has a very long settling time?
The answer is dependent on the definition used for settling time : this is generally taken to mean the time to settle within a given percentage of the steady state value.
To settle precisely to the steady state value, a critically damped system gives the fastest response. An overdamped system will take longer to get there; an underdamped system will get there sooner but overshoot, and take longer overall to reach steady state. (Highly underdamped : it will pass through the steady state point many times)
But to settle within 1% (or 0.1% or 10%) of the steady state value, you can improve on the critically damped settling time by underdamping to such an extent that the first (and worst) overshoot is just within your permitted tolerance.
It still takes longer to reach the actual steady state value; but it is within tolerance faster than a critically damped system. So there is no real contradiction here.
Generally, for an underdamped 2nd order system the 2% settling time is approximately \$\frac{4}{\zeta\:\omega_n}\$. But relating this ROT to, say, a RLC circuit is not intuitive since changing one component value will affect both \$\zeta\$ and \$\omega_n\$.