How to treat the DC in calculate the time constant in RCL circuit having DC voltage source?
I thought \$V_S\$ will act like short circuit


This is why you're confused.

In this circuit, we don't have any control over \$V_S\$ - it is a constant, unchanging, DC voltage. However, we do have control over the switch in the circuit. When, we close the switch, the circuit splits into two separate pieces.

The left side of the circuit becomes:







simulate this circuit &ndash; Schematic created using CircuitLab

Since this circuit has nothing to do with the inductor, we can ignore it.

The right side of the circuit is what your teacher drew, so we can simply analyze that on its own. Note that \$V_S\$ hasn't changed - instead, we've totally taken it out of the circuit.