If $x$, $y$, and $z$ are real numbers for which? Then what is $xyz$

Solve the following system:
{x + y - z = -8 | (equation 1)
x - y + z = 18 | (equation 2)
-x + y + z = 30 | (equation 3)

 

Subtract equation 1 from equation 2:
{x + y - z = -8 | (equation 1)
0 x - 2 y + 2 z = 26 | (equation 2)
-x + y + z = 30 | (equation 3)

 

Divide equation 2 by 2:
{x + y - z = -8 | (equation 1)
0 x - y + z = 13 | (equation 2)
-x + y + z = 30 | (equation 3)

 

Add equation 1 to equation 3:
{x + y - z = -8 | (equation 1)
0 x - y + z = 13 | (equation 2)
0 x+2 y+0 z = 22 | (equation 3)

 

Divide equation 3 by 2:
{x + y - z = -8 | (equation 1)
0 x - y + z = 13 | (equation 2)
0 x+y+0 z = 11 | (equation 3)

 

Add equation 2 to equation 3:
{x + y - z = -8 | (equation 1)
0 x - y + z = 13 | (equation 2)
0 x+0 y+z = 24 | (equation 3)

 

Subtract equation 3 from equation 2:
{x + y - z = -8 | (equation 1)
0 x - y+0 z = -11 | (equation 2)
0 x+0 y+z = 24 | (equation 3)

 

Multiply equation 2 by -1:
{x + y - z = -8 | (equation 1)
0 x+y+0 z = 11 | (equation 2)
0 x+0 y+z = 24 | (equation 3)

 

Subtract equation 2 from equation 1:
{x + 0 y - z = -19 | (equation 1)
0 x+y+0 z = 11 | (equation 2)
0 x+0 y+z = 24 | (equation 3)

 

Add equation 3 to equation 1:
{x+0 y+0 z = 5 | (equation 1)
0 x+y+0 z = 11 | (equation 2)
0 x+0 y+z = 24 | (equation 3)

x = 5,     y = 11,      z = 24          xyz =5  x  11  x  24 =1,320