To find the distance of the plane \( P \) from the origin, we will follow these steps:
### Step 1: Identify the equations of the lines
The given line \( L_1 \) is represented as:
\[
\frac{x}{2} = \frac{y-1}{2} = \frac{z-1}{1}
\]
This can be expressed in parametric form:
\[
x = 2\lambda_1, \quad y = 2\lambda_1 + 1, \quad z = \lambda_1 + 1
\]
The projection of this line onto the plane \( P \) is given by:
\[
\frac{x}{1} = \frac{y-1}{1} = \frac{z-1}{-1}
\]
This can also be expressed in parametric form:
\[
x = \lambda_2, \quad y = \lambda_2 + 1, \quad z = -\lambda_2 + 1
\]
### Step 2: Find points on the lines
Let’s denote a point \( A \) on line \( L_1 \) as:
\[
A(2\lambda_1, 2\lambda_1 + 1, \lambda_1 + 1)
\]
And a corresponding point \( A' \) on line \( L_2 \) as:
\[
A'(\lambda_2, \lambda_2 + 1, -\lambda_2 + 1)
\]
### Step 3: Determine the direction ratios of the segment \( AA' \)
The vector \( \overrightarrow{AA'} \) can be calculated as:
\[
\overrightarrow{AA'} = (2\lambda_1 - \lambda_2, (2\lambda_1 + 1) - (\lambda_2 + 1), (\lambda_1 + 1) - (-\lambda_2 + 1))
\]
Simplifying this gives:
\[
\overrightarrow{AA'} = (2\lambda_1 - \lambda_2, 2\lambda_1 - \lambda_2, \lambda_1 + \lambda_2)
\]
### Step 4: Identify the normal vector of the plane
Since the line \( L_2 \) lies on the plane \( P \), the direction ratios of \( L_2 \) are:
\[
(1, 1, -1)
\]
The normal vector \( \mathbf{n} \) of the plane \( P \) can be expressed as proportional to \( \overrightarrow{AA'} \):
\[
\mathbf{n} = (2\lambda_1 - \lambda_2, 2\lambda_1 - \lambda_2, \lambda_1 + \lambda_2)
\]
### Step 5: Set up the dot product condition
The dot product of the normal vector \( \mathbf{n} \) and the direction ratios of \( L_2 \) must equal zero:
\[
(2\lambda_1 - \lambda_2) \cdot 1 + (2\lambda_1 - \lambda_2) \cdot 1 + (\lambda_1 + \lambda_2) \cdot (-1) = 0
\]
This simplifies to:
\[
2(2\lambda_1 - \lambda_2) + (\lambda_1 + \lambda_2)(-1) = 0
\]
\[
4\lambda_1 - 2\lambda_2 - \lambda_1 - \lambda_2 = 0
\]
\[
3\lambda_1 - 3\lambda_2 = 0 \implies \lambda_1 = \lambda_2
\]
### Step 6: Substitute \( \lambda_2 \) back
Let \( \lambda_2 = \lambda_1 = \lambda \). Then the coordinates of points \( A \) and \( A' \) become:
\[
A(2\lambda, 2\lambda + 1, \lambda + 1)
\]
\[
A'(\lambda, \lambda + 1, -\lambda + 1)
\]
### Step 7: Find the equation of the plane
Using point \( A' \) and normal vector \( \mathbf{n} \):
\[
\text{Equation of the plane: } (x - \lambda)(1) + (y - (\lambda + 1))(1) + (z - (-\lambda + 1))(-1) = 0
\]
This simplifies to:
\[
x + y + 2z - 3 = 0
\]
### Step 8: Calculate the distance from the origin to the plane
The distance \( d \) from the origin \( (0, 0, 0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[
d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
Substituting \( A = 1, B = 1, C = 2, D = -3 \):
\[
d = \frac{|1 \cdot 0 + 1 \cdot 0 + 2 \cdot 0 - 3|}{\sqrt{1^2 + 1^2 + 2^2}} = \frac{3}{\sqrt{6}} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2}
\]
### Final Result
Thus, the distance of the plane \( P \) from the origin is:
\[
\frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2}
\]
