The lines `(x-a+b)/(alpha-delta)=(y-a)/alpha=(z-a-d)/(alpha+delta),`
`(x-b+c)/(beta-gamma)=(y-b)/beta=(z-a-d)/(beta+gamma)`
are coplanar, and the equation to the plane in which they lie is

Options

A. `x+y+z=0`
B. `x-y+z=0`
C. `x-2y+z=0`
D. `x+y+z=0`

Answer

To find the equation of the plane in which the given lines are coplanar, we can follow these steps:

### Step 1: Write down the equations of the lines
The equations of the lines are given as:
1. \(\frac{x-a+b}{\alpha-\delta} = \frac{y-a}{\alpha} = \frac{z-a-d}{\alpha+\delta}\)
2. \(\frac{x-b+c}{\beta-\gamma} = \frac{y-b}{\beta} = \frac{z-a-d}{\beta+\gamma}\)

### Step 2: Convert the lines into parametric form
From the first line, we can express the coordinates \(x\), \(y\), and \(z\) in terms of a parameter \(t\):
- \(x = a - b + t(\alpha - \delta)\)
- \(y = a + t\alpha\)
- \(z = a + d + t(\alpha + \delta)\)

From the second line, we can express the coordinates \(x\), \(y\), and \(z\) in terms of another parameter \(s\):
- \(x = b - c + s(\beta - \gamma)\)
- \(y = b + s\beta\)
- \(z = a + d + s(\beta + \gamma)\)

### Step 3: Set up the determinant for coplanarity
To check if the lines are coplanar, we can use the determinant condition. The determinant is set up as follows:

\[ \begin{vmatrix} x - (a - b) & \alpha - \delta & \beta - \gamma \\ y - a & \alpha & \beta \\ z - (a + d) & \alpha + \delta & \beta + \gamma \end{vmatrix} = 0 \]

### Step 4: Simplify the determinant
We will simplify the determinant step by step. We can perform column operations to simplify it.

1. Add the first column to the third column.
2. Subtract twice the second column from the first column.

After performing these operations, the determinant simplifies to:

\[ \begin{vmatrix} x + z - 2y & 0 & 0 \\ 0 & \alpha - \gamma & \beta - \gamma \\ 0 & \alpha + \delta & \beta + \gamma \end{vmatrix} = 0 \]

### Step 5: Calculate the determinant
The determinant simplifies to:

\[ (x + z - 2y) \cdot \begin{vmatrix} \alpha - \gamma & \beta - \gamma \\ \alpha + \delta & \beta + \gamma \end{vmatrix} = 0 \]

Calculating the 2x2 determinant gives:

\[ (\alpha - \gamma)(\beta + \gamma) - (\beta - \gamma)(\alpha + \delta) \]

### Step 6: Set the equation to zero
Setting the determinant equal to zero gives us:

\[ (x + z - 2y) \cdot K = 0 \]

Where \(K\) is the result of the 2x2 determinant. For the lines to be coplanar, either \(K = 0\) or \(x + z - 2y = 0\).

### Step 7: Write the equation of the plane
Thus, the equation of the plane in which the lines lie is:

\[ x + z - 2y = 0 \]

Rearranging gives:

\[ x - 2y + z = 0 \]

### Final Answer
The required equation of the plane is:

\[ \boxed{x - 2y + z = 0} \]