Can drag be neglected, although the object is in motion? I was researching more about drag and why two things fall at the same time, and I came across Nasa's website (and quite a few more website) which said that, When drag is equal to weight, there is no net external force on the object, and the acceleration becomes zero. The object then falls at a constant velocity. The link to the page is here: https://www.grc.nasa.gov/www/k-12/airplane/falling.html
My question is that, is that information actually true? One reason that says that it is true, is that two objects fall at the same time if left at the same height and same time. This is due to drag becoming equal to weight and then net force is $0$ so $0$ acceleration (according to Nasa's website). But, if you have seen a space probe re-entering earth's atmosphere, you would have probably seen that the front part of the probe sort of catches on fire. Isn't that because of the high drag force, which causes the fire? If you would again go with Nasa, drag should have long ago become equal to weight and been canceled out and it should have not at all caught of fire right? Could someone explain if that information is right or not?
 A: If drag is equal to weight (and there are no other forces), then yes the acceleration will be zero.  For objects falling through the atmosphere, this is commonly referred to as terminal velocity.

One reason that says that it is true, is that two objects fall at the same time if left at the same height and same time. This is due to drag becoming equal to weight and then net force is 0 so 0 acceleration (according to Nasa's website).

But the weight and the drag may be very different.  A marble and a feather do not both fall at the same rate.

If you would again go with Nasa, drag should have long ago become equal to weight and been canceled out and it should have not at all caught of fire right?

No, drag forces depend on the density of the atmosphere, which varies with altitude.  As an object descends, drag increases very quickly, much faster than the object can decelerate.

Does that mean that even after drag being equal to weight, drag increases?

If the fluid it falls through is not constant density, then yes the drag can change until the object reaches a new equilibrium speed.
Imagine an object thrown out of an airplane over the ocean.  After a few seconds it reaches the equilibrium speed where drag and weight are equal.  It falls at (roughly) this speed until it hits the water.  At this point drag increases!  The increased drag slows the object.  Once the object has slowed sufficiently, drag once again equals weight (minus buoyancy) and the object sinks at a slower speed than it did through the air.
The density changes for a reentering spacecraft are not as great, but because of the high speed, the forces are huge.  Rather than smacking the surface of the ocean, the reentering craft strikes the increasing density of the atmosphere.
A: A falling object initially speeds up. The drag force, $$D=\frac12 C_dA\rho v^2,$$ thus gradually increases. At some point the increasing drag force becomes equal to the close-to-constant weight, $w=mg$. Then whichever speed the object has reached as this moment doesn't change anymore - it stays constant from here on. This is called the terminal speed. So yes, the NASA quote is correct.
If the drag force quickly becomes equal to the weight, then a lower terminal speed has been achieved because the object accelerated for a shorter time. So, due to varying drag forces, different objects will reach different terminal speeds.

*

*A parachute, for instance, has a large frontal area, $A$, and a shape that effectively "catches" the air, causing a high drag coefficient $C_d$. The drag force is thus higher and becomes equal to the weight of parachute-plus-person at a much lower speed, $v$.


*To your example, an incoming comet typically enters Earth's atmosphere at a fairly high speed, $v$. When drag starts to have an effect, the enormous speed causes an almost immediately enormous drag force* (note that the speeds is squared in the drag formula). This is why its front start burning with the enormous fluid friction it experiences. While burning at the front, it simultaneously is slowed down rapidly. Soon, the drag force is low enough to equal its weight, and then constant (terminal) speed has been reached.
These two scenarios - a parachute and a comet; something dropped from rest and something entering at high speed - illustrate the two different approaches to terminal speed: either the object starts at a lower-than-terminal-speed and speeds up until the drag has increased enough to balance out the weight, or the object starts at a higher-than-terminal-speed and slows down until the drag has reduced enough to balance out the weight.
In the former case, you see speeding up until constant speed, in the latter you see slowing down until constant speed. Only in the latter case can you see burning at the surface.

* Note that the air density, $\rho$, is lower (the air is thinner) high up and increases as the objects come closer to the ground. This would increase the drag force gradually, meaning that the drag force would become larger than the weight. The object will thus decelerate a bit and slow down a bit until they again balance out. This happens gradually. In other words: the terminal speed changes as the object falls - this is clear from the terminal speed formula which includes the air density:
$$\small{v_\text{terminal}=\sqrt{\frac{2mg}{\rho A C_d}}.}$$
The density quickly becomes fairly even and constant when closer to the ground, so the low-altitude skydiver might not experience such changing terminal speed but will indeed fall at a practically constant terminal speed.**
** In fact the gravitational acceleration, $g$, changes as well with altitude, which would also change the terminal speed gradually. But out atmosphere is not that thick (compared to the size of the planet and the extend of the change in $g$), so the density change happens much faster and is much larger than the $g$-change through the atmosphere, making the $g$-change often practically irrelevant, although it might be included in some delicate calculations.
