Step 1: Understanding the Concept:
Four points are coplanar if the volume of the tetrahedron formed by them is zero. Equivalently, the three vectors originating from one of the points to the other three must lie in the same plane, which means their scalar triple product (or the determinant of their components) must be zero.
Step 2: Key Formula or Approach:
1. Create three vectors from a common point, say D: $\vec{DA}, \vec{DB}, \vec{DC}$.
2. Condition for coplanarity: $[\vec{DA} \quad \vec{DB} \quad \vec{DC}] = 0$, which translates to a determinant being zero.
Step 3: Detailed Explanation:
Let's find the vectors starting from point D$(1, 1, 1)$:
\[ \vec{DA} = A - D = \langle 2-x-1, 2-1, 2-1 \rangle = \langle 1-x, 1, 1 \rangle \]
\[ \vec{DB} = B - D = \langle 2-1, 2-y-1, 2-1 \rangle = \langle 1, 1-y, 1 \rangle \]
\[ \vec{DC} = C - D = \langle 2-1, 2-1, 2-z-1 \rangle = \langle 1, 1, 1-z \rangle \]
For these vectors to be coplanar, their determinant must be zero:
\[ \begin{vmatrix} 1-x & 1 & 1
1 & 1-y & 1
1 & 1 & 1-z \end{vmatrix} = 0 \]
Expand the determinant along the first row:
\[ (1-x)[(1-y)(1-z) - (1)(1)] - 1[(1)(1-z) - (1)(1)] + 1[(1)(1) - (1)(1-y)] = 0 \]
Simplify the terms inside the brackets:
\[ (1-x)[1 - z - y + yz - 1] - 1[1 - z - 1] + 1[1 - 1 + y] = 0 \]
\[ (1-x)[yz - y - z] - 1[-z] + 1[y] = 0 \]
Expand the first product:
\[ (yz - y - z) - x(yz - y - z) + z + y = 0 \]
\[ yz - y - z - xyz + xy + xz + z + y = 0 \]
Notice that $-y + y = 0$ and $-z + z = 0$, so they cancel out:
\[ yz - xyz + xy + xz = 0 \]
Rearrange the terms:
\[ xy + yz + zx = xyz \]
Divide the entire equation by $xyz$ (assuming $x, y, z \neq 0$ to find a valid locus form matching the options):
\[ \frac{xy}{xyz} + \frac{yz}{xyz} + \frac{zx}{xyz} = \frac{xyz}{xyz} \]
\[ \frac{1}{z} + \frac{1}{x} + \frac{1}{y} = 1 \]
Rearranging in alphabetical order:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 \]
Step 4: Final Answer:
The locus is $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$.