If the points `(2-x, 2, 2), (2, 2-y, 2) and (2, 2, 2-z)` are coplanar then prove `2/x+2/y+2/z=1`

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. `1/(1-x)+1/(1-y)+1/(1-z)=2`

Explanation

To prove that the points \((2-x, 2, 2)\), \((2, 2-y, 2)\), and \((2, 2, 2-z)\) are coplanar and to show that \( \frac{2}{x} + \frac{2}{y} + \frac{2}{z} = 1 \), we can follow these steps:

### Step 1: Set up the determinant for coplanarity
The points are coplanar if the determinant of the matrix formed by their coordinates is zero. We can express this as:

\[ \begin{vmatrix} 2-x & 2 & 2 \\ 2 & 2-y & 2 \\ 2 & 2 & 2-z \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant
We will calculate the determinant using the method of cofactor expansion. The determinant can be expanded as follows:

\[ D = (2-x) \begin{vmatrix} 2-y & 2 \\ 2 & 2-z \end{vmatrix} - 2 \begin{vmatrix} 2 & 2 \\ 2 & 2-z \end{vmatrix} + 2 \begin{vmatrix} 2 & 2-y \\ 2 & 2 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
Now, we will calculate each of the 2x2 determinants:

1. For \(\begin{vmatrix} 2-y & 2 \\ 2 & 2-z \end{vmatrix}\):
\[ = (2-y)(2-z) - 2 \cdot 2 = (4 - 2z - 2y + yz) - 4 = yz - 2y - 2z \]

2. For \(\begin{vmatrix} 2 & 2 \\ 2 & 2-z \end{vmatrix}\):
\[ = 2(2-z) - 2 \cdot 2 = 4 - 2z - 4 = -2z \]

3. For \(\begin{vmatrix} 2 & 2-y \\ 2 & 2 \end{vmatrix}\):
\[ = 2 \cdot 2 - 2(2-y) = 4 - (4 - 2y) = 2y \]

### Step 4: Substitute back into the determinant
Now substituting these back into the determinant \(D\):

\[ D = (2-x)(yz - 2y - 2z) - 2(-2z) + 2(2y) \]
\[ = (2-x)(yz - 2y - 2z) + 4z + 4y \]

### Step 5: Set the determinant to zero
Setting the determinant \(D\) to zero gives us:

\[ (2-x)(yz - 2y - 2z) + 4z + 4y = 0 \]

### Step 6: Expand and simplify
Now we expand and simplify the equation:

\[ (2-x)yz - 2(2-x)y - 2(2-x)z + 4z + 4y = 0 \]

This leads to:

\[ 2yz - xyz - 4y + 2xy + 4z - 2xz = 0 \]

### Step 7: Rearranging the equation
Rearranging gives us:

\[ 2yz + 2xy + 2xz = xyz \]

### Step 8: Divide by xyz
Dividing the entire equation by \(xyz\) (assuming \(x, y, z \neq 0\)) gives:

\[ \frac{2}{x} + \frac{2}{y} + \frac{2}{z} = 1 \]

### Conclusion
Thus, we have proved that if the points are coplanar, then:

\[ \frac{2}{x} + \frac{2}{y} + \frac{2}{z} = 1 \]