Question: A regular tetrahedron has vertices at $(1, 1, 1)$, $(1, -1, -1)$, $(-1, 1, -1)$, and $(a, b, c)$ with integer coordinates. Find $(a, b, c)$.
["A regular tetrahedron has vertices at $(1, 1, 1)$, $(1, -1, -1)$, $(-1, 1, -1)$, and $(a, b, c)$ with integer coordinates. Find $(a, b, c)$. \nIn the world of geometry, a regular tetrahedron stands out as a shape defined by four equilateral triangle faces—symmetrical, precise, and deeply studied. With three known vertices plotted in 3D space,Finding the fourth with purely integer coordinates feels like a puzzle gaining traction beyond academic circles. This question matters now more than ever, as curiosity about spatial patterns thrives in online communities focused on math, design, and digital art—especially in the U.S., where visual literacy and spatial reasoning are increasingly vital. Blooming interest in geometry, structures, and design systems fuels ongoing discussion, making this a timely and relevant topic.", "This question is gaining attention not for sensationalism but for its elegant algebraic and geometric challenge. A regular tetrahedron’s vertices share equal pairwise distances, and given three integer-coordinate points, solving for the fourth becomes both a math puzzle and a test of spatial logic—but without resorting to creative leaps or hidden assumptions. Users seek clarity, accuracy, and trustworthy solutions online—especially when dealing with subtle coordinate geometry in a world where precision matters in architecture, product design, and digital modeling.", "So, what’s the actual solution? How do we find $(a, b, c)$, the fourth vertex with integer coordinates?", "To approach this systematically, begin by analyzing the known points: $A(1, 1, 1)$, $B(1, -1, -1)$, $C(-1, 1, -1)$. First, compute all pairwise distances to confirm these form an equilateral triangle—necessary for forming a regular tetrahedron. The distance between $A$ and $B$ is:", "\[\n\sqrt{(1-1)^2 + (1+1)^2 + (1+1)^2} = \sqrt{0 + 4 + 4} = \sqrt{8}\n\]", "Similarly, distances $AC$ and $BC$ also equal $\sqrt{8}$, confirming an equilateral base in 3D space. Let $D(a, b, c)$ be the fourth vertex. For the full tetrahedron to be regular, the distance from $D$ to each of $A$, $B$, $C$ must also be $\sqrt{8}$, and all four vertices must lie symmetrically.", "Without loss of generality, exploiting geometric symmetry and vector algebra reveals that the centroid of the equilateral base lies at:", "\[\nG = \left( \frac{1 + 1 - 1 + a}{4}, \frac{1 - 1 + 1 + b}{4}, \frac{1 - 1 - 1 + c}{4} "]
