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-5z-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 given lines, we will follow these steps:

### Step 1: Identify the direction ratios of the lines
The two lines are given in symmetric form:
1. Line 1: \((x-1)/2 = (y+1)/(-1) = z/3\)
- Direction ratios \(b_1 = (2, -1, 3)\)

2. Line 2: \(x/(-1) = (y-2)/3 = (z+1)/(-1)\)
- Direction ratios \(b_2 = (-1, 3, -1)\)

### Step 2: Find a point on each line
For Line 1, when \(t = 0\):
- \(x = 1\), \(y = -1\), \(z = 0\) → Point \(A(1, -1, 0)\)

For Line 2, when \(s = 0\):
- \(x = 0\), \(y = 2\), \(z = -1\) → Point \(B(0, 2, -1)\)

### Step 3: Find the normal vector to the plane
The normal vector \(n\) to the plane can be found using the cross product of the direction vectors \(b_1\) and \(b_2\).

\[ b_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}, \quad b_2 = \begin{pmatrix} -1 \\ 3 \\ -1 \end{pmatrix} \]

The cross product \(n = b_1 \times b_2\) is calculated as follows:

\[ n = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ -1 & 3 & -1 \end{vmatrix} \]

Calculating the determinant:

\[ n = \hat{i} \begin{vmatrix} -1 & 3 \\ 3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 3 \\ -1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ -1 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} -1 & 3 \\ 3 & -1 \end{vmatrix} = (-1)(-1) - (3)(3) = 1 - 9 = -8\)
2. \(\begin{vmatrix} 2 & 3 \\ -1 & -1 \end{vmatrix} = (2)(-1) - (3)(-1) = -2 + 3 = 1\)
3. \(\begin{vmatrix} 2 & -1 \\ -1 & 3 \end{vmatrix} = (2)(3) - (-1)(-1) = 6 - 1 = 5\)

Thus, we have:

\[ n = -8 \hat{i} - 1 \hat{j} + 5 \hat{k} = (-8, -1, 5) \]

### Step 4: Use the point-normal form of the plane equation
Using point \(A(1, -1, 0)\) and normal vector \(n(-8, -1, 5)\), the equation of the plane can be written as:

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

Expanding this:

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

Rearranging gives:

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

Multiplying through by -1 to simplify:

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

### Final Equation of the Plane
The equation of the plane is:

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