Let
n ----> the first odd integer
n+2 ----> the second consecutive odd integer
n+4---> the third consecutive odd integer
so
The equation that represents this problem is
n+(n+2)+(n+4)=-321
3n+6=-321
solve for n
3n=-321-6
3n=-327
n=-109
n+2=-109+2=-107
n+4=-109+4=--105
therefore
The numbers are
-109,-107 and -105