To solve the problem step by step, we will follow the outlined approach to find the value of \( \lambda \).
### Step 1: Identify the Points
We have three points through which the plane \( P \) passes:
- \( A(2, 1, 0) \)
- \( B(\lambda - 1, 1, 1) \)
- \( C(\lambda, 0, 1) \)
We also have the point \( R(2, 1, 6) \) and its image \( Q(6, 5, -2) \).
### Step 2: Find the Midpoint \( M \) of \( R \) and \( Q \)
The midpoint \( M \) of points \( R \) and \( Q \) can be calculated using the midpoint formula:
\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)
\]
Substituting the coordinates of \( R \) and \( Q \):
\[
M = \left( \frac{2 + 6}{2}, \frac{1 + 5}{2}, \frac{6 - 2}{2} \right) = \left( 4, 3, 2 \right)
\]
### Step 3: Find the Equation of the Plane \( P \)
The equation of a plane can be determined using the determinant method. The general form for the equation of the plane through points \( A, B, C \) is given by:
\[
\begin{vmatrix}
x - 2 & y - 1 & z - 0 \\
(\lambda - 1) - 2 & 1 - 1 & 1 - 0 \\
\lambda - 2 & 0 - 1 & 1 - 0
\end{vmatrix} = 0
\]
Calculating the determinant:
\[
\begin{vmatrix}
x - 2 & y - 1 & z \\
\lambda - 3 & 0 & 1 \\
\lambda - 2 & -1 & 1
\end{vmatrix} = 0
\]
Expanding this determinant:
\[
(x - 2) \begin{vmatrix}
0 & 1 \\
-1 & 1
\end{vmatrix} - (y - 1) \begin{vmatrix}
\lambda - 3 & 1 \\
\lambda - 2 & 1
\end{vmatrix} + z \begin{vmatrix}
\lambda - 3 & 0 \\
\lambda - 2 & -1
\end{vmatrix} = 0
\]
Calculating the smaller determinants:
1. \( \begin{vmatrix} 0 & 1 \\ -1 & 1 \end{vmatrix} = 0 \cdot 1 - 1 \cdot (-1) = 1 \)
2. \( \begin{vmatrix} \lambda - 3 & 1 \\ \lambda - 2 & 1 \end{vmatrix} = (\lambda - 3) \cdot 1 - 1 \cdot (\lambda - 2) = \lambda - 3 - \lambda + 2 = -1 \)
3. \( \begin{vmatrix} \lambda - 3 & 0 \\ \lambda - 2 & -1 \end{vmatrix} = (\lambda - 3)(-1) - 0 = -\lambda + 3 \)
Substituting back into the equation:
\[
(x - 2)(1) - (y - 1)(-1) + z(-\lambda + 3) = 0
\]
This simplifies to:
\[
x - 2 + y - 1 - z(\lambda - 3) = 0
\]
Rearranging gives us the equation of the plane:
\[
x + y - z(\lambda - 3) - 3 = 0
\]
### Step 4: Substitute Point \( M \) into the Plane Equation
Since \( M(4, 3, 2) \) lies on the plane, substitute \( x = 4 \), \( y = 3 \), and \( z = 2 \):
\[
4 + 3 - 2(\lambda - 3) - 3 = 0
\]
This simplifies to:
\[
4 + 3 - 3 - 2\lambda + 6 = 0
\]
\[
10 - 2\lambda = 0
\]
Solving for \( \lambda \):
\[
2\lambda = 10 \implies \lambda = 5
\]
### Final Answer
Thus, the value of \( \lambda \) is:
\[
\lambda = 5
\]
