According to the question,
5a5=7a7
5[a+(5−1)d]=7[a+(7−1)d]
5(a+4d)=7(a+6d)
5a+20d=7a+42d
2a+22d=0
a+11d=0
a+(12−1)d=0
a12=0
Therefore, the 12th term is 0.