The coplanar points `A,B,C,D` are `(2-x,2,2),(2,2-y,2),(2,2,2-z)` and `(1,1,1)` respectively then

Options

A. `1/x+1/y+1/z=1`
B. `x+y+z=1`
C. `1/(1-x)+1/(1-y)+1/(1-z)=1`
D. none of these

Explanation

To determine the relationship between the variables \( x \), \( y \), and \( z \) given the coplanar points \( A, B, C, D \), we can follow these steps:

### Step 1: Define the Points
The points are given as:
- \( A = (2-x, 2, 2) \)
- \( B = (2, 2-y, 2) \)
- \( C = (2, 2, 2-z) \)
- \( D = (1, 1, 1) \)

### Step 2: Find Vectors
To analyze the coplanarity, we need to find vectors from point \( A \) to points \( B \), \( C \), and \( D \):
- Vector \( \vec{AB} = B - A = (2 - (2-x), (2-y) - 2, 2 - 2) = (x, -y, 0) \)
- Vector \( \vec{AC} = C - A = (2 - (2-x), 2 - 2, (2-z) - 2) = (x, 0, -z) \)
- Vector \( \vec{AD} = D - A = (1 - (2-x), 1 - 2, 1 - 2) = (x - 1, -1, -1) \)

### Step 3: Set Up the Determinant
The points are coplanar if the volume of the parallelepiped formed by these vectors is zero. This can be expressed using the determinant of a matrix formed by these vectors:
\[ \begin{vmatrix} x & -y & 0 \\ x & 0 & -z \\ x - 1 & -1 & -1 \end{vmatrix} = 0 \]

### Step 4: Calculate the Determinant
We will compute the determinant:
\[ \text{Det} = x \begin{vmatrix} 0 & -z \\ -1 & -1 \end{vmatrix} - (-y) \begin{vmatrix} x & -z \\ x - 1 & -1 \end{vmatrix} + 0 \begin{vmatrix} x & 0 \\ x - 1 & -1 \end{vmatrix} \]

Calculating the first determinant:
\[ \begin{vmatrix} 0 & -z \\ -1 & -1 \end{vmatrix} = 0 \cdot (-1) - (-z)(-1) = -z \]

Calculating the second determinant:
\[ \begin{vmatrix} x & -z \\ x - 1 & -1 \end{vmatrix} = x(-1) - (-z)(x - 1) = -x + z(x - 1) = zx - z - x \]

Putting it all together:
\[ \text{Det} = x(-z) + y(zx - z - x) = -xz + y(zx - z - x) \]

### Step 5: Set the Determinant to Zero
Setting the determinant to zero gives us:
\[ -xz + y(zx - z - x) = 0 \]

### Step 6: Simplify the Equation
Expanding and rearranging:
\[ -xz + yzx - yz - yx = 0 \]
This can be rearranged to:
\[ y(zx - y - x) = xz \]

### Step 7: Final Relationship
From the above equation, we can derive the relationship:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 \]