We have to solve the system of equations:

y = -x - 3 ...(1)

x = -z - 2...(2)

y(x+z) = 2 - xz...(3)

substitute x = -z - 2 in (1)

=> y = (z + 2) - 3 = z - 1

Substitute x = -z - 2 and y = z - 1 in (3)

=> (z - 1)(-z - 2 + z) = 2 - (-z - 2)z

=> (z - 1)(-2) = 2 + z^2 + 2z

=> -2z + 2 = 2 + z^2 + 2z

=> z^2 + 4z = 0

=> z(z + 4) = 0

=> z = 0 and z = -4

x = -z - 2 => x = -2 and x = 2

y = z - 1 => y = -1 and y = -5

**The solution for the system is (x, y, z) = (2 , -5 , -4) and (-2 , -1 , 0)**

We'll re-write the 1st,the 2nd and the 3rd equations, moving all the variables to the left side:

x + y = -3 (1)

x + z = -2 (2)

xy+xz+yz = 2 (3)

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Now, we'll multiply the 1st and the 2nd equations:

(x+y)(x+z) = (-3)*(-2)

We'll remove the brackets:

x^2 + xz + xy + yz = 6

We notice that the sum xz + xy + yz can be substituted by the 3rd equation:

x^2 + 2= 6

x^2 = 6-2

x^2 = 4

x1 = 2 and x2 = -2

We'll substitute x1 in the 1st equation:

2 + y = -3

y = -2-3

y = -5

We'll substitute x1 in the 2nd equation:

2 +z = -2

z = -4

We'll substitute x2 in the 1st equation:

-2 + y = -3

y = -1

We'll substitute x2 in the 2nd equation:

-2 + z = -2

z = 0

**The solutions of the system of equations are represented by the pairs of coordinates(x,y,z): (2 , -5 , -4) and (-2 ,-1 , 0).**