We use here a coordinate system with the constant covariant metric:
1 −½ −½
−½ 1 −½
−½ −½ 1
Or g_ij = (3/2)δ_ij - ½. (See this about similar coordinate systems.)

In this system the following are unit vectors ±(1, 0, 0), ±(0, 1, 0), ±(0, 0, 1), ±(0, 1, 1), ±(1, 0, 1), ±(1, 1, 0). Just now I see that the length of (1, 1, 1) is 0 (recall that Length² = g_ijxⁱx^j) and that distresses me. Indeed this shows that the eigenvalues of g are 0, 3/2 and 3/2.

Adopt plain coordinates, x, y, z. The basis vectors used above in plain coordinates are:
(a, b, b), (b, a, b), (b, b, a) where a² + 2b² = 1 (since their lengths are 1) and 2ab+b² = −½ (since the dot product of any two of them is −½).
a² = 1 − 2b²
2(√(1 − 2b²))b+b² = −½
2(√(1 − 2b²))b = −½ − b²
√(1 − 2b²) = (−½ − b²)/(2b)
1 − 2b² = (1/(4b) + b/2)²
b² − 2b⁴ = (¼ + b²/2)²
b² − 2b⁴ = 1/16 + b²/4 + b⁴/4
16b² − 32b⁴ = 1 + 4b² + 4b⁴
36b⁴ − 12b² + 1 = 0
b² = (12±√(144 − 144))/72 = 1/6
b = −√(1/6) = −0.408248290463863
a = √(⅔) = 0.816496580927726

The point which is at (1, 1, 1) in oblique coordinates is at (c, c, c) in plain coordinates where c = a + 2b = 0. Oops. The puzzle pieces are not shaped as I thought!

This is obvious in retrospect. I required that the basis vectors for our oblique system be 120° from each other. In 3D three such vectors must lie in the same plane and thus cannot span the space.