If the points `A(3-x, 3, 3), B(3, 3-y, 3), C(3, 3-y, 3) and C(3, 3, 3-z)D(2, 2, 2)` are coplanar, then `(1)/(x)+(1)/(y)+(1)/(z)` is equal to

Options

A. `-1`
B. 1
C. 3
D. 5

Answer

To solve the problem of determining the value of \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\) given that the points \(A(3-x, 3, 3)\), \(B(3, 3-y, 3)\), \(C(3, 3, 3-z)\), and \(D(2, 2, 2)\) are coplanar, we will follow these steps:

### Step 1: Find the position vectors of points A, B, C, and D
The position vectors of the points are:
- \( \vec{A} = (3-x, 3, 3) \)
- \( \vec{B} = (3, 3-y, 3) \)
- \( \vec{C} = (3, 3, 3-z) \)
- \( \vec{D} = (2, 2, 2) \)

### Step 2: Calculate the vectors \(\vec{DA}\), \(\vec{DB}\), and \(\vec{DC}\)
We calculate the vectors from point D to points A, B, and C:
- \( \vec{DA} = \vec{A} - \vec{D} = (3-x-2, 3-2, 3-2) = (1-x, 1, 1) \)
- \( \vec{DB} = \vec{B} - \vec{D} = (3-2, 3-y-2, 3-2) = (1, 1-y, 1) \)
- \( \vec{DC} = \vec{C} - \vec{D} = (3-2, 3-2, 3-z-2) = (1, 1, 1-z) \)

### Step 3: Set up the determinant for coplanarity
The points A, B, C, and D are coplanar if the scalar triple product of the vectors \(\vec{DA}\), \(\vec{DB}\), and \(\vec{DC}\) is zero. This can be expressed as the determinant:

\[ \begin{vmatrix} 1-x & 1 & 1 \\ 1 & 1-y & 1 \\ 1 & 1 & 1-z \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Expanding the determinant, we have:

\[ (1-x) \begin{vmatrix} 1-y & 1 \\ 1 & 1-z \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & 1-z \end{vmatrix} + 1 \begin{vmatrix} 1 & 1-y \\ 1 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 1-y & 1 \\ 1 & 1-z \end{vmatrix} = (1-y)(1-z) - 1 = 1 - y - z + yz - 1 = yz - y - z \)

2. \( \begin{vmatrix} 1 & 1 \\ 1 & 1-z \end{vmatrix} = 1(1-z) - 1 = 1 - z - 1 = -z \)

3. \( \begin{vmatrix} 1 & 1-y \\ 1 & 1 \end{vmatrix} = 1(1) - 1(1-y) = 1 - (1-y) = y \)

Putting it all together:

\[ (1-x)(yz - y - z) + z + y = 0 \]

### Step 5: Expand and simplify the equation
Expanding gives:

\[ (1-x)yz - (1-x)y - (1-x)z + z + y = 0 \]

Rearranging terms leads to:

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

### Step 6: Rearranging the equation
We can rearrange this to find:

\[ xyz = xy + xz + yz \]

### Step 7: Divide through by \(xyz\)
Dividing by \(xyz\) gives:

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

Thus, we find:

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

### Final Answer
The value of \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\) is:

\[ \boxed{1} \]