Hi suraj,
The 78L05 is a protected device and hard to destroy.
The minimum input voltage is 8.5 volt. A series diode will solve your problem. The diode only needs to be able to carry 52 mA (load current + 78L05 current.
The voltage drop across even a small diode should be around 1 volt maximum and its dissipation only 55 milli watt. Most small signal diodes will dissipate at least 150 milli watt.
The use of the diode will reduce the power supply input to the 78L05 from 12 to say 11 volt. As this is above the minimum of 8.5 volt, then you are OK. The reduced input voltage to the 78L05 will reduce its internal dissipation, which is good.
1N4148 is usually available.
Hope this helps.