To solve this system of equations, we can start by setting each pair of equations equal to each other:

1) x/y + z + 1 = y/x + z + 1
2) y/x + z + 1 = z/x + y - 2
3) z/x + y - 2 = x + y + z

From equation 1, we have:
x/y = y/x

Cross multiplying, we get:
x^2 = y^2

Taking the square root of both sides, we get:
x = ±y

Now, we can substitute x = y and x = -y into the other equations to find the values of z:

1) x/y + z + 1 = y/x + z + 1
If x = y:
1 + z + 1 = 1/z + z + 1
2z = 1/z
2z^2 = 1
z^2 = 1/2
z = ±√(1/2)

2) y/x + z + 1 = z/x + y - 2
If x = y:
1 + z + 1 = 1/z + z - 2
2z = 1/z - 3
2z^2 = 1 - 3z
2z^2 + 3z - 1 = 0
Solving this quadratic equation gives us the values of z.

3) z/x + y - 2 = x + y + z
If x = y:
1 + z - 2 = 1 + z + z
z - 1 = 2z
z = -1

Therefore, the solutions to the system of equations are:
1) x = y, z = ±√(1/2)
2) x = y, z = z from solving the quadratic equation
3) x = y, z = -1