Zener breakdown occurs due to heavily doped diodes.In this process electrons crosses the barrier from 

Zener breakdown occurs in heavily doped p-n junctions. The heavy doping makes the depletion layer extremely thin. So that, carriers cannot accelerate enough to cause ionization. Thus, current will increase in reverse bias only due to reverse breakdown voltage.

Avalanche breakdown is caused by impact ionization of electron-hole pairs. A very little current flows under reverse bias conditions and depletion region increases. The electric field in the depletion region of a diode can be very high. Electron/holes that enter the depletion region undergo a tremendous acceleration. As these accelerated carriers collide with the atoms, they can knock electrons from their bonds, creating additional electron/hole pairs and thus additional current. As these secondary carriers are swept into the depletion region, they too are accelerated and the process repeats itself.

The avalanche breakdown in p-n junction is due to cumulative effect of conduction band electron. In reverse bias, due to applied electric field, electrons acquire enough energy to free more electron bound to the atom. The abundant number of electron-hole pairs are created for conduction, this is cumulative effect.

The reverse breakdown voltage depends on doping of the diode. Hence, in the Zener diode the heavy doping of its p-n junction is done. The depletion region formed in the diode is very thin ($<1\ m$) and the reverse bias voltage of about $5\ V$ which is less than ordinary diode.

When the reverse bias breakdown voltage is exceeded, a conventional diode is subject to high current due to avalanche breakdown. Avalanche breakdown occurs in reverse bias when the applied voltage is high enough, the free electron may move fast enough to knock other electrons free, creating more free-electron-hole pairs (i.e., more charge carriers), increasing the current. Thus, the current flow in a Zener diode is mainly due to collision generated charge carriers

We know that zener diode works on the reverse bias. When the reverse bias is equal to the break-down voltage, the voltage across the zener remains almost constant and the current increases rapidly.

Zener diode is used to supply constant voltage in voltage regulator circuit hence option (b) is correct.

When the reverse voltage across a diode is very large, the valance electrons become free due to applied high electric field and get enough acceleration to make other electrons free, thus create a lot of electron-hole pairs in a short time. As the number of charge carriers increases, current also increases.

Zener diode is a p-n junction diode working in the breakdown region. It is used as a voltage regulator/stabilizer to provide a constant voltage from a source whose voltage may fluctuate over a wide range.

 A voltage regulator circuit can be designed using a zener diode to maintain a constant DC output voltage across the load in spite of variations in the input voltage or changes in the load current. 

True. A voltage regulator circuit can be designed using a zener diode to maintain a constant DC output voltage across the load in spite of variations in the input voltage or changes in the load current. 

current increases as the resistance decreases, as zener current is inversely proportional to the series resistance

For Zener Diodes, the breakdown voltage decreases with increasing doping concentration.

The zener voltage will be a constant only when it is operating in the breakdown region

Consider applying an external voltage to the junction with the the negative terminal connected to the P type material and the positive terminal connected to the N type material. For the P-N junction this would require injecting electrons into the P type material. These electrons would recombine with holes and therefore further deplete the majority charge carriers in the P type semiconductor. 

Zener diode is used as a voltage regulator device because the voltage across the zener diode in reverse bias is constant after reaching certain value i.e. breakdown voltage.

As the current in the zener diode increases, the voltage across the diode remains constant. This voltage will be equal to the zener breakdown voltage

The load current is independent of the source voltage variations and hence remains a constant

The load current is given by $I _L=\dfrac{V _z}{R _L}$ and the zener current is given by $I _z=I _s-I _L$. Thus, if the load resistance decreases, load current increases. This in turn will reduce the zener current