To find the distance of the point \( (1, 1, 1) \) from the plane that passes through the point \( (-1, -2, -1) \) and whose normal is perpendicular to both given lines, we can follow these steps:
### Step 1: Identify the direction ratios of the lines
The two lines are given in symmetric form:
1. Line 1: \(\frac{x+1}{3} = \frac{y+2}{1} = \frac{z+1}{2}\)
2. Line 2: \(\frac{x-2}{1} = \frac{y+2}{2} = \frac{z-3}{3}\)
From these equations, we can extract the direction ratios:
- For Line 1, the direction ratios are \( (3, 1, 2) \).
- For Line 2, the direction ratios are \( (1, 2, 3) \).
### Step 2: Find the normal vector to the plane
The normal vector to the plane can be found by taking the cross product of the direction ratios of the two lines.
Let:
\[
\mathbf{A} = (3, 1, 2) \quad \text{and} \quad \mathbf{B} = (1, 2, 3)
\]
The cross product \(\mathbf{A} \times \mathbf{B}\) is calculated as follows:
\[
\mathbf{A} \times \mathbf{B} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 1 & 2 \\
1 & 2 & 3
\end{vmatrix}
\]
Calculating the determinant:
\[
= \mathbf{i} \left( 1 \cdot 3 - 2 \cdot 2 \right) - \mathbf{j} \left( 3 \cdot 3 - 2 \cdot 1 \right) + \mathbf{k} \left( 3 \cdot 2 - 1 \cdot 1 \right)
\]
\[
= \mathbf{i} (3 - 4) - \mathbf{j} (9 - 2) + \mathbf{k} (6 - 1)
\]
\[
= -\mathbf{i} - 7\mathbf{j} + 5\mathbf{k}
\]
Thus, the normal vector is \( (-1, -7, 5) \).
### Step 3: Write the equation of the plane
The equation of a plane can be expressed as:
\[
A(x - x_0) + B(y - y_0) + C(z - z_0) = 0
\]
Where \( (x_0, y_0, z_0) \) is a point on the plane, and \( (A, B, C) \) are the components of the normal vector.
Using the point \( (-1, -2, -1) \) and the normal vector \( (-1, -7, 5) \):
\[
-1(x + 1) - 7(y + 2) + 5(z + 1) = 0
\]
Expanding this gives:
\[
-x - 1 - 7y - 14 + 5z + 5 = 0
\]
Rearranging:
\[
-x - 7y + 5z - 10 = 0 \quad \Rightarrow \quad x + 7y - 5z + 10 = 0
\]
### Step 4: Find the distance from the point to the plane
The formula for the distance \( D \) from a point \( (x_1, y_1, z_1) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[
D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
Here, \( (x_1, y_1, z_1) = (1, 1, 1) \) and the coefficients from the plane equation \( A = 1, B = 7, C = -5, D = 10 \).
Calculating the numerator:
\[
|1 \cdot 1 + 7 \cdot 1 - 5 \cdot 1 + 10| = |1 + 7 - 5 + 10| = |13| = 13
\]
Calculating the denominator:
\[
\sqrt{1^2 + 7^2 + (-5)^2} = \sqrt{1 + 49 + 25} = \sqrt{75}
\]
Thus, the distance \( D \) is:
\[
D = \frac{13}{\sqrt{75}} = \frac{13}{5\sqrt{3}} = \frac{13\sqrt{3}}{15}
\]
### Final Answer
The distance of the point \( (1, 1, 1) \) from the plane is \( \frac{13\sqrt{3}}{15} \).
