Let x be the first number, y the second one, and z the third one, then we can set the following system of equations:

Adding the last equation to the first one we get:

Now, adding the first two equations and substituting x=2 we get:

Finally, substituting x=, z=-2 in the first equation and solving for y we get:

Answer: The first number is 2, the second one is 3 and the third one is -2.