### Instruction: A certain regular tetrahedron has three of its vertices at the points $(0,1,2),$ $(4,2,1),$ and $(3,1,5).$  Find the coordinates of the fourth vertex, given that they are also all integers.### Response: A regular tetrahedron is a solid with four faces, each of which is an equilateral triangle.
That means that the four vertices are all equidistant from each other.
Let $(x,y,z)$ be the coordinates of the fourth vertex.
Then the distance from $(x,y,z)$ to $(0,1,2)$ is the same as the distance from $(0,1,2)$ to $(4,2,1)$.
I can use the distance formula to write this as an equation:
\[ \sqrt{(x-0)^2 + (y-1)^2 + (z-2)^2} = \sqrt{(4-0)^2 + (2-1)^2 + (1-2)^2} \]
Squaring both sides and simplifying, I get
\[ x^2 + (y-1)^2 + (z-2)^2 = 18 \]
Similarly, the distance from $(x,y,z)$ to $(3,1,5)$ is the same as the distance from $(3,1,5)$ to $(4,2,1)$.
Another equation is
\[ \sqrt{(x-3)^2 + (y-1)^2 + (z-5)^2} = \sqrt{(4-3)^2 + (2-1)^2 + (1-5)^2} \]
Squaring and simplifying, I get
\[ (x-3)^2 + (y-1)^2 + (z-5)^2 = 18 \]
Now I have two equations in three variables.
I need one more equation to be able to solve for $(x,y,z)$.
I notice that the vectors $\overrightarrow{(0,1,2)(4,2,1)}$ and $\overrightarrow{(0,1,2)(3,1,5)}$ are both edges of the tetrahedron, and they are orthogonal to each other, since the angle between them is $60^\circ$.
That means that their dot product is zero.
I can use this fact to write another equation:
\[ \overrightarrow{(0,1,2)(4,2,1)} \cdot \overrightarrow{(0,1,2)(3,1,5)} = 0 \]
Expanding the dot product, I get
\[ (4-0)(3-0) + (2-1)(1-1) + (1-2)(5-2) = 0 \]
Simplifying, I get
\[ 12 - 8 = 0 \]
This is a contradiction.
This means that the given vertices do not form a regular tetrahedron, and there is no fourth vertex with integer coordinates that satisfies the conditions.