= Solution
The three known vertices have pairwise squared distances $8$, so the <regular tetrahedron> must have every squared edge length equal to $8$. Comparing the squared distance from $d$ to $b$ with that from $d$ to $a$ gives
$$
|d-b|^2-|d-a|^2
=4+(x+1)^2-(x-1)^2=4+4x.
$$
Equality forces $x=-1$. Similarly,
$$
|d-c|^2-|d-a|^2=4+(y+1)^2-(y-1)^2=4+4y,
$$
so $y=-1$. The resulting point has squared distance $8$ from each of the other three vertices. Therefore
$$
\boxed{x=y=-1,\qquad d=(1,-1,-1).}
$$
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