To determine 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 can follow these steps:
### Step 1: Find the vectors \( \overrightarrow{DA}, \overrightarrow{DB}, \overrightarrow{DC} \)
The vectors from point \(D\) to points \(A\), \(B\), and \(C\) are calculated as follows:
\[
\overrightarrow{DA} = A - D = (3-x - 2, 3 - 2, 3 - 2) = (1-x, 1, 1)
\]
\[
\overrightarrow{DB} = B - D = (3 - 2, 3-y - 2, 3 - 2) = (1, 1-y, 1)
\]
\[
\overrightarrow{DC} = C - D = (3 - 2, 3 - 2, 3-z - 2) = (1, 1, 1-z)
\]
### Step 2: Set up the determinant for coplanarity
The points \(A\), \(B\), \(C\), and \(D\) are coplanar if the scalar triple product of the vectors \(\overrightarrow{DA}\), \(\overrightarrow{DB}\), and \(\overrightarrow{DC}\) is zero. This can be expressed using the determinant:
\[
\begin{vmatrix}
1-x & 1 & 1 \\
1 & 1-y & 1 \\
1 & 1 & 1-z
\end{vmatrix} = 0
\]
### Step 3: Calculate the determinant
Calculating the determinant, we expand it:
\[
= (1-x) \begin{vmatrix}
1-y & 1 \\
1 & 1-z
\end{vmatrix} - 1 \begin{vmatrix}
1 & 1 \\
1 & 1-z
\end{vmatrix} + 1 \begin{vmatrix}
1 & 1-y \\
1 & 1
\end{vmatrix}
\]
Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} 1-y & 1 \\ 1 & 1-z \end{vmatrix} = (1-y)(1-z) - 1 = 1 - y - z + yz - 1 = -y - z + yz\)
2. \(\begin{vmatrix} 1 & 1 \\ 1 & 1-z \end{vmatrix} = 1(1-z) - 1 = 1 - z - 1 = -z\)
3. \(\begin{vmatrix} 1 & 1-y \\ 1 & 1 \end{vmatrix} = 1(1) - 1(1-y) = 1 - (1-y) = y\)
Substituting these back into the determinant gives:
\[
(1-x)(-y - z + yz) - (-z) + y = 0
\]
### Step 4: Simplify the equation
Expanding this gives:
\[
-(1-x)(y + z - yz) + z + y = 0
\]
Distributing \(-(1-x)\):
\[
-(y + z - yz) + x(y + z - yz) + z + y = 0
\]
Rearranging gives:
\[
x(y + z - yz) + z + y - (y + z - yz) = 0
\]
This simplifies to:
\[
x(y + z - yz) + yz = 0
\]
### Step 5: Solve for \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\)
From the coplanarity condition, we can express the relationship:
\[
x(yz - y - z) = 0
\]
This implies \(x + y + z = xyz\) when rearranged. Dividing through by \(xyz\) gives:
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1
\]
### Final Answer
Thus, we find that:
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1
\]
