Find three consecutive odd integers such that eight more than the sum of the first two is equal to eleven less then three times the third

A+b+8 = 11-3c
a+b+3c = 11-8
a+b+3c = 3

a = b-1
c = b+1

b-1+b+3*(b+1) = 3
2b-b+3b+3 = 3
4b = 3-3
4b = 0
b = 0
a = b-1
a = 0-1
a = -1
c = b+1
c = 0+1
c = 1

a = -1, b = 0, c = 1

10,000 hope i ! : )