To find the distance between the two planes \( P_1 \) and \( P_2 \) given the equations of two intersecting lines in plane \( P_1 \) and the equation of plane \( P_2 \), we will follow these steps:
### Step 1: Write the equations of the lines
The equations of the two intersecting lines are given as:
1. \( \frac{x-1}{1} = \frac{y-3}{2} = \frac{z-4}{3} \)
2. \( \frac{x-1}{2} = \frac{y-3}{3} = \frac{z-4}{1} \)
### Step 2: Identify the direction ratios of the lines
From the equations, we can extract the direction ratios:
- For the first line, the direction ratios are \( (1, 2, 3) \).
- For the second line, the direction ratios are \( (2, 3, 1) \).
### Step 3: Find the normal vector to the plane \( P_1 \)
The normal vector \( \mathbf{n} \) to the plane can be found using the cross product of the direction vectors of the two lines. Let:
- \( \mathbf{r_1} = (1, 2, 3) \)
- \( \mathbf{r_2} = (2, 3, 1) \)
The cross product \( \mathbf{n} = \mathbf{r_1} \times \mathbf{r_2} \) is calculated as follows:
\[
\mathbf{n} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 3 \\
2 & 3 & 1
\end{vmatrix}
\]
Calculating the determinant:
\[
\mathbf{n} = \mathbf{i}(2 \cdot 1 - 3 \cdot 3) - \mathbf{j}(1 \cdot 1 - 3 \cdot 2) + \mathbf{k}(1 \cdot 3 - 2 \cdot 2)
\]
\[
= \mathbf{i}(2 - 9) - \mathbf{j}(1 - 6) + \mathbf{k}(3 - 4)
\]
\[
= -7\mathbf{i} + 5\mathbf{j} - 1\mathbf{k}
\]
Thus, the normal vector \( \mathbf{n} = (-7, 5, -1) \).
### Step 4: Find a point on the plane \( P_1 \)
To find a point on the plane, we can take the intersection point of the two lines. Setting \( t = 0 \) in the first line gives the point \( (1, 3, 4) \).
### Step 5: Write the equation of plane \( P_1 \)
Using the normal vector \( (-7, 5, -1) \) and the point \( (1, 3, 4) \), the equation of the plane can be written as:
\[
-7(x - 1) + 5(y - 3) - 1(z - 4) = 0
\]
Expanding this:
\[
-7x + 7 + 5y - 15 - z + 4 = 0
\]
\[
-7x + 5y - z - 4 = 0
\]
Rearranging gives:
\[
7x - 5y + z + 4 = 0
\]
### Step 6: Write the equation of plane \( P_2 \)
The equation of plane \( P_2 \) is given as:
\[
7x - 5y + z - 6 = 0
\]
### Step 7: Calculate the distance between planes \( P_1 \) and \( P_2 \)
The distance \( d \) between two parallel planes of the form \( Ax + By + Cz + D_1 = 0 \) and \( Ax + By + Cz + D_2 = 0 \) is given by:
\[
d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}}
\]
Here, \( D_1 = 4 \) and \( D_2 = -6 \). Thus:
\[
d = \frac{|(-6) - 4|}{\sqrt{7^2 + (-5)^2 + 1^2}} = \frac{|-10|}{\sqrt{49 + 25 + 1}} = \frac{10}{\sqrt{75}} = \frac{10}{5\sqrt{3}} = \frac{2}{\sqrt{3}}
\]
### Final Answer
The distance between the planes \( P_1 \) and \( P_2 \) is \( \frac{2}{\sqrt{3}} \).
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