Equation of plane containing the two lines
`(x-1)/2 = (y+1)/-1 = z/3` and `x/2 = (y-2)/-1 = (z+1)/3` is

Options

A. 8x+y-5z=7
B. 9x-5y + 6z+2=0
C. 3x 4y + 7z=5
D. 9x+5y + 6z=38

Answer

To find the equation of the plane that contains the two given lines, we will follow these steps:

### Step 1: Identify Points and Direction Ratios of the Lines

**Line 1:**
Given the equation \((x-1)/2 = (y+1)/-1 = z/3\), we can extract:
- A point on the line: \(P_1(1, -1, 0)\)
- Direction ratios: \(d_1(2, -1, 3)\)

**Line 2:**
Given the equation \(x/2 = (y-2)/-1 = (z+1)/3\), we can extract:
- A point on the line: \(P_2(0, 2, -1)\)
- Direction ratios: \(d_2(2, -1, 3)\)

### Step 2: Write the General Equation of the Plane

The general 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 direction ratios normal to the plane.

Using point \(P_1(1, -1, 0)\):
\[ a(x - 1) + b(y + 1) + c(z - 0) = 0 \]
This simplifies to:
\[ a(x - 1) + b(y + 1) + cz = 0 \]

### Step 3: Find the Condition for the Plane to Contain the Lines

Since the plane contains both lines, the direction ratios of the lines must satisfy the plane's equation.

For line 1:
The direction ratios \(d_1 = (2, -1, 3)\) must satisfy:
\[ 2a - b + 3c = 0 \quad \text{(1)}\]

For line 2:
The direction ratios \(d_2 = (2, -1, 3)\) must satisfy:
\[ 2a - b + 3c = 0 \quad \text{(2)}\]

### Step 4: Substitute Point \(P_2(0, 2, -1)\) into the Plane Equation

Since the plane must also contain point \(P_2\):
Substituting into the plane equation:
\[ a(0 - 1) + b(2 + 1) + c(-1 - 0) = 0 \]
This simplifies to:
\[ -a + 3b - c = 0 \quad \text{(3)}\]

### Step 5: Solve the System of Equations

Now we have a system of equations:
1. \(2a - b + 3c = 0\)
2. \(-a + 3b - c = 0\)

From equation (1):
\[ b = 2a + 3c \]

Substituting \(b\) in equation (3):
\[-a + 3(2a + 3c) - c = 0\]
\[-a + 6a + 9c - c = 0\]
\[5a + 8c = 0\]
Thus, \(a = -\frac{8}{5}c\).

### Step 6: Choose a Value for \(c\) and Find \(a\) and \(b\)

Let \(c = 5\):
Then,
\[ a = -8 \]
\[ b = 2(-8) + 3(5) = -16 + 15 = -1 \]

### Step 7: Write the Equation of the Plane

Now substituting \(a, b, c\) back into the plane equation:
\[ -8(x - 1) - 1(y + 1) + 5z = 0 \]
Expanding this gives:
\[ -8x + 8 - y - 1 + 5z = 0 \]
Which simplifies to:
\[ 8x + y - 5z = 7 \]

### Final Answer

The equation of the plane is:
\[ 8x + y - 5z = 7 \]