Divide the first equation by 2 to get:
x + 2y + 3z = 0
4x + 5y + 6z = 3
7x + 8y + 9z = 6
Add -4 times equation 1 to equation 2 and add -7 times equation 1 to equation 3:
x + 2y + 3z = 0
-3y -6z = 3
-6y -12z = 6
Add -2 times equation 2 to equation 3:
x + 2y + 3z = 0
-3y - 6z = 3
0 = 0
From equation 2, we have y = -2z -1
Substitute for y in equation 1: x + 2(-2z - 1) + 3z = 0. So, x = z + 2
We have infinitely many solutions of the form:
x = z + 2
y = -2z - 1
z = any real number
For example, if z = 0, then we get the solution (2, -1, 0). Letting z = 1 yields the solution (3, -3, 1), etc.
