To find the equation of the plane passing through the point of intersection of the given lines and perpendicular to another line, we can follow these steps:
### Step 1: Identify the direction vectors of the given lines
The first line is given by:
\[
\frac{x-1}{3} = \frac{y-2}{1} = \frac{z-3}{2}
\]
From this, we can extract the direction vector \( \mathbf{d_1} = 3\mathbf{i} + 1\mathbf{j} + 2\mathbf{k} \).
The second line is given by:
\[
\frac{x-3}{1} = \frac{y-1}{2} = \frac{z-2}{3}
\]
From this, we can extract the direction vector \( \mathbf{d_2} = 1\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \).
### Step 2: Find the normal vector of the plane
The plane is perpendicular to the line given by:
\[
\frac{x+5}{2} = \frac{y-3}{3} = \frac{z+1}{1}
\]
From this, we can extract the direction vector \( \mathbf{b} = 2\mathbf{i} + 3\mathbf{j} + 1\mathbf{k} \). This vector will be the normal vector \( \mathbf{n} \) of the required plane.
### Step 3: Find the point of intersection of the two lines
Let the point of intersection of the two lines be \( (x_1, y_1, z_1) \).
For the first line, we can set:
\[
x_1 = 3\lambda + 1, \quad y_1 = \lambda + 2, \quad z_1 = 2\lambda + 3
\]
Let’s assume this is equal to \( \mu \) for the second line:
\[
x_1 = \mu + 3, \quad y_1 = 2\mu + 1, \quad z_1 = 3\mu + 2
\]
### Step 4: Set up equations based on the point of intersection
From the equations of the two lines, we can equate the coordinates:
1. \( 3\lambda + 1 = \mu + 3 \)
2. \( \lambda + 2 = 2\mu + 1 \)
3. \( 2\lambda + 3 = 3\mu + 2 \)
### Step 5: Solve the system of equations
From the first equation:
\[
3\lambda - \mu = 2 \quad \text{(Equation 1)}
\]
From the second equation:
\[
\lambda - 2\mu = -1 \quad \text{(Equation 2)}
\]
From the third equation:
\[
2\lambda - 3\mu = -1 \quad \text{(Equation 3)}
\]
Now, we can solve these equations.
From Equation 1, we can express \( \mu \):
\[
\mu = 3\lambda - 2
\]
Substituting \( \mu \) into Equation 2:
\[
\lambda - 2(3\lambda - 2) = -1
\]
\[
\lambda - 6\lambda + 4 = -1
\]
\[
-5\lambda = -5 \implies \lambda = 1
\]
Now substituting \( \lambda = 1 \) back into the expression for \( \mu \):
\[
\mu = 3(1) - 2 = 1
\]
### Step 6: Find the coordinates of the point of intersection
Substituting \( \lambda = 1 \) into the equations for \( (x_1, y_1, z_1) \):
\[
x_1 = 3(1) + 1 = 4, \quad y_1 = 1 + 2 = 3, \quad z_1 = 2(1) + 3 = 5
\]
Thus, the point of intersection is \( P(4, 3, 5) \).
### Step 7: Write the equation of the plane
The equation of the plane can be given by:
\[
\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}
\]
Where \( \mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \) and \( \mathbf{a} = 4\mathbf{i} + 3\mathbf{j} + 5\mathbf{k} \).
Calculating \( \mathbf{a} \cdot \mathbf{n} \):
\[
\mathbf{n} = 2\mathbf{i} + 3\mathbf{j} + 1\mathbf{k}
\]
\[
\mathbf{a} \cdot \mathbf{n} = 4(2) + 3(3) + 5(1) = 8 + 9 + 5 = 22
\]
Thus, the equation of the plane is:
\[
2x + 3y + z = 22
\]
### Final Answer
The equation of the plane is:
\[
2x + 3y + z - 22 = 0
\]
