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 solve the problem of finding the value of \(7\lambda\) given that the lines

\[ \frac{x-1}{2} = \frac{y}{-1} = \frac{z}{2} \]

and

\[ x - y + z - 2 = 0 = \lambda x + 3z + 5 \]

are coplanar, we will follow these steps:

### Step 1: Identify the Direction Ratios and Point of the First Line
The first line can be expressed in parametric form. From the equation

\[ \frac{x-1}{2} = \frac{y}{-1} = \frac{z}{2} = t, \]

we can write the parametric equations as:

\[ x = 2t + 1, \quad y = -t, \quad z = 2t. \]

The direction ratios of the first line are \( (2, -1, 2) \) and a point on the line is \( (1, 0, 0) \).

### Step 2: Identify the Normal Vector of the Second Line
The second line is given by the equation

\[ \lambda x + 3z + 5 = 0. \]

Rearranging gives us:

\[ \lambda x + 3z = -5. \]

This can be interpreted as a plane equation where the normal vector \( \mathbf{n_2} \) is \( (\lambda, 0, 3) \).

### Step 3: Find the Normal Vector of the Plane
The plane defined by the equation \( x - y + z - 2 = 0 \) has a normal vector \( \mathbf{n_1} = (1, -1, 1) \).

### Step 4: Use the Condition for Coplanarity
For the lines to be coplanar, the scalar triple product of the direction ratios of the lines and the normal vector of the plane must be zero. We can set up the following matrix:

\[ \begin{vmatrix} 2 & -1 & 2 \\ \lambda & 0 & 3 \\ 1 & -1 & 1 \end{vmatrix} = 0. \]

### Step 5: Calculate the Determinant
Calculating the determinant, we have:

\[ \begin{vmatrix} 2 & -1 & 2 \\ \lambda & 0 & 3 \\ 1 & -1 & 1 \end{vmatrix} = 2 \begin{vmatrix} 0 & 3 \\ -1 & 1 \end{vmatrix} + 1 \begin{vmatrix} \lambda & 3 \\ 1 & 1 \end{vmatrix} + 2 \begin{vmatrix} \lambda & 0 \\ 1 & -1 \end{vmatrix}. \]

Calculating each of these 2x2 determinants:

1. \( 2(0 \cdot 1 - 3 \cdot (-1)) = 2(3) = 6 \).
2. \( 1(\lambda \cdot 1 - 3 \cdot 1) = \lambda - 3 \).
3. \( 2(\lambda \cdot (-1) - 0 \cdot 1) = -2\lambda \).

Putting it all together:

\[ 6 + (\lambda - 3) - 2\lambda = 0. \]

### Step 6: Solve for \(\lambda\)
Combining like terms:

\[ 6 - 3 - \lambda = 0 \implies 3 - \lambda = 0 \implies \lambda = 3. \]

### Step 7: Find \(7\lambda\)
Now, we can find \(7\lambda\):

\[ 7\lambda = 7 \cdot 3 = 21. \]

Thus, the final answer is:

\[ \boxed{21}. \]