We use here a coordinate system with the constant covariant metric:
 1 −½ −½ −½ 1 −½ −½ −½ 1
Or gij = (3/2)δij - 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 = gijxixj) 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.