Let,the resistance of each of the three lamp is r,
Now, lamp X and Y are in parallel combination,so their net resistance is (r*r)/(r+r) =r/2
They are again in series with lamp Z, so net resistance of the circuit is r+ r/2 =(3r)/2
If the voltage of the battery is V ,then current flowing through the circuit is V/((3r)/2)=(2V)/(3r)
So,current flowing through lamp Z is (2V)/(3r)
Now,potential drop across lamp X and Y is the same as they are in parallel combination, So the value is (V - (2V)/(3r)*r)=V/3
So,current flowing through lamp Y is (V/3) /r =V/(3r)
So,initially, more current is flowing through lamp Z than X,so it was glowing more.
when X is no more, lamp Z and Y comes in series combination,so net resistance is r+r=2r hence,current flowing through each of them is V/(2r)
So, this value of current is higher w.r.t initial value of current flowing through lamp Y so it becomes glower, but for Z this value has decreased than previous,so its brightness decreases.
