Let the first odd integer be x

The second will be (x+2)

Third odd integer = (x+4)

Given sum is = 141

So,

x+(x+2)+(x+4)=141

3x+6=141

3x=141-6

3x=135

x=135/3

x=45

x=45, (x+2)=47, (x+4)=49

45, 47, and 49 are the three consecutive odd integers whose sum is 141.