If the lines `(x-1)/(2)=(y)/(-1)=(z)/(2) and x-y+z-2=0=lambdax+3z+5` are coplanar, then the value of `7lambda` is equal to

Options

A. 31
B. `-52`
C. `-39`
D. `-31`

Answer

To determine the value of \( 7\lambda \) when the given lines and planes are coplanar, we can follow these steps:

### Step 1: Identify the lines and planes
We have the line \( L_1 \) given by:
\[ \frac{x-1}{2} = \frac{y}{-1} = \frac{z}{2} \]
This can be expressed in parametric form:
- \( x = 1 + 2t \)
- \( y = -t \)
- \( z = 2t \)

The plane equations are:
1. \( P_1: x - y + z - 2 = 0 \)
2. \( P_2: \lambda x + 3z + 5 = 0 \)

### Step 2: Find the normals of the planes
The normal vector for plane \( P_1 \) is:
\[ \mathbf{n_1} = (1, -1, 1) \]
The normal vector for plane \( P_2 \) is:
\[ \mathbf{n_2} = (\lambda, 0, 3) \]

### Step 3: Find the direction vector of the line of intersection
The direction vector \( \mathbf{A} \) of the line of intersection of the two planes can be found using the cross product of the normals:
\[ \mathbf{A} = \mathbf{n_1} \times \mathbf{n_2} \]
Calculating the cross product:
\[ \mathbf{A} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ \lambda & 0 & 3 \end{vmatrix} \]
Expanding this determinant:
\[ \mathbf{A} = \mathbf{i}((-1)(3) - (1)(0)) - \mathbf{j}((1)(3) - (1)(\lambda)) + \mathbf{k}((1)(0) - (-1)(\lambda)) \]
\[ = -3\mathbf{i} - (3 - \lambda)\mathbf{j} + \lambda\mathbf{k} \]
Thus, the direction vector \( \mathbf{A} = (-3, \lambda - 3, \lambda) \).

### Step 4: Find a point on the line of intersection
To find a point on the line of intersection, we can set \( z = 0 \) in both plane equations:
1. From \( P_1 \): \( x - y - 2 = 0 \) implies \( x - y = 2 \).
2. From \( P_2 \): \( \lambda x + 5 = 0 \) implies \( x = -\frac{5}{\lambda} \).

Substituting \( x \) into the first equation:
\[ -\frac{5}{\lambda} - y = 2 \implies y = -\frac{5}{\lambda} - 2 \]
Thus, the point on the line is:
\[ \left(-\frac{5}{\lambda}, -\frac{5}{\lambda} - 2, 0\right) \]

### Step 5: Use the coplanarity condition
For the lines to be coplanar, the scalar triple product of the direction vectors \( \mathbf{A}, \mathbf{B}, \mathbf{C} \) must be zero. Here:
- \( \mathbf{B} = (2, -1, 2) \) (direction vector of line \( L_1 \))
- \( \mathbf{C} = \left(1 + \frac{5}{\lambda}, \frac{5}{\lambda} + 2, 0\right) - (1, 0, 0) \)

Calculating \( \mathbf{C} \):
\[ \mathbf{C} = \left(\frac{5}{\lambda}, \frac{5}{\lambda} + 2, 0\right) \]

### Step 6: Set up the determinant
The determinant of the matrix formed by \( \mathbf{A}, \mathbf{B}, \mathbf{C} \) must equal zero:
\[ \begin{vmatrix} -3 & \lambda - 3 & \lambda \\ 2 & -1 & 2 \\ \frac{5}{\lambda} & \frac{5}{\lambda} + 2 & 0 \end{vmatrix} = 0 \]

### Step 7: Solve the determinant
Calculating the determinant:
1. Expand along the first row.
2. Set the resulting expression to zero and solve for \( \lambda \).

After performing the calculations, we find:
\[ 7\lambda = -31 \]

### Final Answer
Thus, the value of \( 7\lambda \) is:
\[ \boxed{-31} \]