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

Options

A. `8x+y-5x-7=0`
B. `8x+y+5z-7=0`
C. `8x-y-5z-7=0`
D. none of these

Explanation

To find the equation of the plane containing the two lines given by the equations:

1. \((x-1)/2 = (y+1)/(-1) = z/3\)
2. \(x/(-1) = (y-2)/3 = (z+1)/(-1)\)

we will follow these steps:

### Step 1: Identify Points on Each Line
From the first line, we can express it in parametric form. Let \( t \) be the parameter:
- \( x = 1 + 2t \)
- \( y = -1 - t \)
- \( z = 3t \)

When \( t = 0 \), we get the point \( A(1, -1, 0) \).

From the second line, let \( s \) be the parameter:
- \( x = -s \)
- \( y = 2 + 3s \)
- \( z = -1 - s \)

When \( s = 0 \), we get the point \( B(0, 2, -1) \).

### Step 2: Find Direction Vectors of the Lines
For the first line, the direction vector \( \mathbf{b_1} \) can be derived from the coefficients of \( t \):
- \( \mathbf{b_1} = \langle 2, -1, 3 \rangle \)

For the second line, the direction vector \( \mathbf{b_2} \) can be derived from the coefficients of \( s \):
- \( \mathbf{b_2} = \langle -1, 3, -1 \rangle \)

### Step 3: Find the Normal Vector of the Plane
The normal vector \( \mathbf{n} \) to the plane can be found using the cross product of the two direction vectors \( \mathbf{b_1} \) and \( \mathbf{b_2} \):

\[ \mathbf{n} = \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ -1 & 3 & -1 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{n} = \mathbf{i}((-1)(-1) - (3)(3)) - \mathbf{j}((2)(-1) - (3)(-1)) + \mathbf{k}((2)(3) - (-1)(-1)) \]

\[ = \mathbf{i}(1 - 9) - \mathbf{j}(-2 + 3) + \mathbf{k}(6 - 1) \]

\[ = -8\mathbf{i} - 1\mathbf{j} + 5\mathbf{k} \]

Thus, the normal vector is \( \mathbf{n} = \langle -8, -1, 5 \rangle \).

### Step 4: Use the Point-Normal Form of the Plane
The equation of the plane can be expressed using the point-normal form:

\[ \mathbf{n} \cdot (\mathbf{r} - \mathbf{a}) = 0 \]

Where \( \mathbf{r} = \langle x, y, z \rangle \) and \( \mathbf{a} = \langle 1, -1, 0 \rangle \).

Substituting in the values:

\[ \langle -8, -1, 5 \rangle \cdot \langle x - 1, y + 1, z - 0 \rangle = 0 \]

This expands to:

\[ -8(x - 1) - 1(y + 1) + 5z = 0 \]

Simplifying gives:

\[ -8x + 8 - y - 1 + 5z = 0 \]

\[ -8x - y + 5z + 7 = 0 \]

Multiplying through by -1 to get a standard form:

\[ 8x + y - 5z - 7 = 0 \]

### Final Answer
Thus, the equation of the plane is:

\[ 8x + y - 5z - 7 = 0 \]