For this type of system of equation was developed a general solution.

for this case a=1, b=2 and c=3.

I will also provide a full solution.

First, I notice that if subtract equations one from another I will end up with the formula of difference of squares. I will try it and see what happens:

x^2 - y^2 -yz + zx = 1 - 2

(x - y)(x + y) + z(x - y) = -1

I will then factorize by taking common factor (x - y) outside the brackets:

(x - y)(x + y + z) = -1 **(1)**

I will repeat that with two other pairs of equations and will get:

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

(y - z)(x + y + z) = -1 **(3)**

I notice that right parts of the equations (1) and (3) are equal to each other so left parts should be equal too:

(x - y)(x + y + z) = (y - z)(x + y + z), or

x - y = y - z **(4)**

The same apply to equations (1)*2 and (2), therefore:

2(x - y) = x - z **(5)**

Next I will try to add all three equations together and see if I can apply the formula of square of difference/sum. (*Please note, this system of equations is not something standard we have ready strategy for. I just keep trying different methods and see what works.*)

x^2 + y^2 + z^2 - yz - zx - xy = 6

I know that formula for difference of two squares looks like that a^2 - 2ab + b^2 = (a - b)^2 so I will multiply both parts of equation by 2:

2x^2 + 2y^2 + 2z^2 - 2yz - 2zx - 2xy = 12

After that I can group my term:

(x^2 - 2xy + y^2) + (x^2 - 2xz + z^2) + (y^2 -2yz - z^2) = 12, or

(x - y)^2 + (x - z)^2 + (y - z)^2 = 12 **(6)**

Now I will substitute (4) and (5) into (6):

(x - y)^2 + 4(x - z)^2 + (x - z)^2 = 12

6(x - y)^2 = 12

(x - y)^2 = 2, or