Let ω be the complex number cos(2π/3) + i sin(2π/3). Then the number of distinct complex numbers z satisfying ||z + 1 ω ω² ωz + ω² 1 ω² 1 z + ω|| = 0 is equal to:

| | z + 1 ω ω² ωz + ω² 1 ω² 1 z + ω | | = 0
Let C1 = z + 1, C2 = ωz + ω², C3 = ω²z + ω. Then the given equation is equivalent to |C1, C2, C3| = 0.
This implies that C1, C2, C3 are linearly dependent.
Hence, there exist scalars a, b, c such that aC1 + bC2 + cC3 = 0, where a, b, c are not all zero.
Then a(z+1) + b(ωz+ω²) + c(ω²z+ω) = 0
(a+bω+cω²)z + a + bω² + cω = 0
For this equation to hold for all z, the coefficients of z and the constant term must be zero.
Therefore, we have the system of equations:
a + bω + cω² = 0
a + bω² + cω = 0
Subtracting the second equation from the first, we get:
b(ω-ω²) + c(ω²-ω) = 0
b(ω-ω²) - c(ω-ω²) = 0
(b-c)(ω-ω²) = 0
Since ω ≠ ω², we have b = c.
Substituting b = c into the first equation, we get:
a + bω + bω² = 0
a + b(ω + ω²) = 0
Since ω + ω² = -1, we have a - b = 0, which means a = b = c.
Therefore, a = b = c = k for some scalar k.
So, we have k(z+1) + k(ωz+ω²) + k(ω²z+ω) = 0
k(z+1+ωz+ω²+ω²z+ω) = 0
k(z(1+ω+ω²) + 1+ω+ω²) = 0
Since 1+ω+ω² = 0, we have 0 = 0, which is always true.
This means that C1, C2, C3 are linearly dependent for any z.
However, for there to be 0 determinant, the vectors must be linearly dependent. This means that any z satisfies this equation. However, the question asks for the number of distinct complex numbers. We have an identity that holds true for any z. Thus there are infinitely many solutions for z. However, if we look at the context of the question, it seems the questioner is looking for a different property which is not clearly expressed. The provided solution and the question statement seem to be incompatible.
The given solution is incomplete and doesn't lead to a numerical answer.