To find the equation of the plane passing through the given parallel lines, we will follow these steps:
### Step 1: Identify Points and Direction Ratios of the Lines
The given lines are:
1. \(\frac{x-3}{3} = \frac{y+4}{2} = \frac{z-1}{1}\)
2. \(\frac{x+1}{3} = \frac{y-2}{2} = z\)
From the first line, we can extract:
- A point on the first line: \(P_1(3, -4, 1)\)
- Direction ratios: \(d_1 = (3, 2, 1)\)
From the second line, we can extract:
- A point on the second line: \(P_2(-1, 2, 0)\)
- Direction ratios: \(d_2 = (3, 2, 1)\)
### Step 2: Write the General Equation of the Plane
The general equation of a plane can be expressed as:
\[
a(x - x_1) + b(y - y_1) + c(z - z_1) = 0
\]
where \((x_1, y_1, z_1)\) is a point on the plane.
Using point \(P_1(3, -4, 1)\):
\[
a(x - 3) + b(y + 4) + c(z - 1) = 0
\]
### Step 3: Substitute the Second Point into the Plane Equation
Since the plane also passes through point \(P_2(-1, 2, 0)\), we substitute these coordinates into the plane equation:
\[
a(-1 - 3) + b(2 + 4) + c(0 - 1) = 0
\]
This simplifies to:
\[
-4a + 6b - c = 0 \quad \text{(Equation 1)}
\]
### Step 4: Use the Direction Ratios of the Lines
Since the lines are parallel, their direction ratios \(d = (3, 2, 1)\) must be perpendicular to the normal vector of the plane \((a, b, c)\). Therefore, we have:
\[
3a + 2b + c = 0 \quad \text{(Equation 2)}
\]
### Step 5: Solve the System of Equations
Now we have two equations:
1. \(-4a + 6b - c = 0\)
2. \(3a + 2b + c = 0\)
We can add these two equations to eliminate \(c\):
\[
(-4a + 6b - c) + (3a + 2b + c) = 0
\]
This simplifies to:
\[
-a + 8b = 0 \implies a = 8b
\]
### Step 6: Substitute \(a\) into One of the Equations
Substituting \(a = 8b\) into Equation 2:
\[
3(8b) + 2b + c = 0 \implies 24b + 2b + c = 0 \implies 26b + c = 0 \implies c = -26b
\]
### Step 7: Express \(a\), \(b\), and \(c\) in Terms of \(b\)
Let \(b = k\) (a constant), then:
\[
a = 8k, \quad b = k, \quad c = -26k
\]
### Step 8: Substitute Back into the Plane Equation
Substituting \(a\), \(b\), and \(c\) back into the plane equation:
\[
8k(x - 3) + k(y + 4) - 26k(z - 1) = 0
\]
Dividing through by \(k\) (assuming \(k \neq 0\)):
\[
8(x - 3) + (y + 4) - 26(z - 1) = 0
\]
Expanding this gives:
\[
8x - 24 + y + 4 - 26z + 26 = 0
\]
Combining like terms:
\[
8x + y - 26z + 6 = 0
\]
### Final Equation of the Plane
Thus, the equation of the plane is:
\[
8x + y - 26z + 6 = 0
\]
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