Solution: Let the fourth vertex be $D = (x, y, z)$. In a regular tetrahedron, all six edges must be equal. First, compute the distance between known points: - AIKO, infinite ways to autonomy.

Title: Solving for Vertex $D = (x, y, z)$ in a Regular Tetrahedron: Equal Edge Lengths Explained

Introduction
Creating a regular tetrahedron in 3D space requires all six edges to be equal in length—this presents a classic geometric challenge. In this article, we explore a key solution approach: solving for the unknown fourth vertex $D = (x, y, z)$ when several vertices are already defined. By computing distances between known points and enforcing uniform edge lengths, we establish equations that determine the precise coordinates of $D$, ensuring symmetry and regularity. Let’s dive into the details.

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Now subtract (1) from (3):
$$
\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{a\sqrt{3}}{2}\right)^2 + z^2 - (x^2 + y^2 + z^2) = 0
$$
Substitute $x = \frac{a}{2}$:
$$
0 - 2\left(\frac{a}{2}\right)\cdot\frac{a}{2} + \frac{a^2}{4} + y^2 - 2y\cdot\frac{a\sqrt{3}}{2} + \frac{3a^2}{4} = 0
\Rightarrow -\frac{a^2}{2} + \frac{a^2}{4} + y^2 - a\sqrt{3}\, y + \frac{3a^2}{4} = 0
\Rightarrow \left(-\frac{1}{2} + \frac{1}{4} + \frac{3}{4}\right)a^2 + y^2 - a\sqrt{3}\, y = 0
\Rightarrow a^2 + y^2 - a\sqrt{3}\, y = 0
$$
Solve the quadratic in $y$:
$$
y^2 - a\sqrt{3}\, y + a^2 = 0
\Rightarrow y = \frac{a\sqrt{3} \pm \sqrt{3a^2 - 4a^2}}{2} = \frac{a\sqrt{3} \pm \sqrt{-a^2}}{2}
$$
Wait—this suggests an error in sign. Rechecking the algebra, the correct expansion yields:
From:
$$
-\frac{a^2}{2} + \frac{a^2}{4} + \frac{3a^2}{4} + y^2 - a\sqrt{3}\, y = 0
\Rightarrow \left(-\frac{1}{2} + 1\right)a^2 + y^2 - a\sqrt{3}\, y = 0
\Rightarrow \frac{1}{2}a^2 + y^2 - a\sqrt{3}\, y = 0
\Rightarrow y^2 - a\sqrt{3}\, y + \frac{1}{2}a^2 = 0
$$
Discriminant:
$$
\Delta = (a\sqrt{3})^2 - 4 \cdot 1 \cdot \frac{1}{2}a^2 = 3a^2 - 2a^2 = a^2 > 0
\Rightarrow y = \frac{a\sqrt{3} \pm a}{2}
$$
Thus:
- $y = \frac{a(\sqrt{3} + 1)}{2}$ or $y = \frac{a(\sqrt{3} - 1)}{2}$

Now substitute $x = \frac{a}{2}$, $y$, and solve for $z$ using equation (1):

From (1):
$$
\left(\frac{a}{2}\right)^2 + y^2 + z^2 = a^2 \Rightarrow \frac{a^2}{4} + y^2 + z^2 = a^2
\Rightarrow z^2 = a^2 - \frac{a^2}{4} - y^2 = \frac{3a^2}{4} - y^2
$$

Using $y = \frac{a(\sqrt{3} - 1)}{2}$:
$$
y^2 = \frac{a^2}{4} \cdot ( (\sqrt{3} - 1)^2 ) = \frac{a^2}{4} (3 - 2\sqrt{3} + 1) = \frac{a^2}{4} (4 - 2\sqrt{3}) = \frac{a^2}{2}(2 - \sqrt{3})
$$
Then:
$$
z^2 = \frac{3a^2}{4} - \frac{a^2}{2}(2 - \sqrt{3}) = a^2\left( \frac{3}{4} - 1 + \frac{\sqrt{3}}{2} \right) = a^2\left( -\frac{1}{4} + \frac{\sqrt{3}}{2} \right) = a^2 \left( \frac{2\sqrt{3} - 1}{4} \right)
$$
Thus:
$$
z = \pm a \sqrt{ \frac{2\sqrt{3} - 1}{4} }
$$