The lines `(x-2)/(1) = (y-3)/(1) =(z-4)/(-k) and (x-3)/(k)=(y-4)/(1) = (z-5)/(1)` are coplanar if the values of k are

Options

A. 0 or -1
B. 1 or -1
C. 0 or -3
D. 3 or -3

Answer

To determine the values of \( k \) for which the given lines are coplanar, we will follow these steps:

### Step 1: Write the equations of the lines in symmetric form
The lines are given as:
1. \(\frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{-k}\)
2. \(\frac{x-3}{k} = \frac{y-4}{1} = \frac{z-5}{1}\)

From these equations, we can identify the points and direction ratios of the lines.

### Step 2: Identify points and direction ratios
For the first line:
- Point \( P_1(2, 3, 4) \)
- Direction ratios \( (1, 1, -k) \)

For the second line:
- Point \( P_2(3, 4, 5) \)
- Direction ratios \( (k, 1, 1) \)

### Step 3: Use the coplanarity condition
The lines are coplanar if the determinant formed by the direction ratios and the vector connecting the two points is zero. The condition for coplanarity can be expressed as:

\[ \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0 \]

Where:
- \( (x_1, y_1, z_1) = (2, 3, 4) \)
- \( (x_2, y_2, z_2) = (3, 4, 5) \)
- Direction ratios \( (a_1, b_1, c_1) = (1, 1, -k) \)
- Direction ratios \( (a_2, b_2, c_2) = (k, 1, 1) \)

### Step 4: Calculate the determinant
First, calculate the differences:
- \( x_2 - x_1 = 3 - 2 = 1 \)
- \( y_2 - y_1 = 4 - 3 = 1 \)
- \( z_2 - z_1 = 5 - 4 = 1 \)

Now, substituting these values into the determinant:

\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & -k \\ k & 1 & 1 \end{vmatrix} = 0 \]

### Step 5: Expand the determinant
Calculating the determinant:

\[ = 1 \cdot \begin{vmatrix} 1 & -k \\ 1 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & -k \\ k & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 \\ k & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & -k \\ 1 & 1 \end{vmatrix} = 1 \cdot 1 - (-k) \cdot 1 = 1 + k \)
2. \( \begin{vmatrix} 1 & -k \\ k & 1 \end{vmatrix} = 1 \cdot 1 - (-k) \cdot k = 1 + k^2 \)
3. \( \begin{vmatrix} 1 & 1 \\ k & 1 \end{vmatrix} = 1 \cdot 1 - 1 \cdot k = 1 - k \)

Putting it all together:

\[ 1(1 + k) - 1(1 + k^2) + 1(1 - k) = 0 \]

### Step 6: Simplify the equation
This simplifies to:

\[ 1 + k - 1 - k^2 + 1 - k = 0 \]

Combining like terms:

\[ -k^2 + 1 = 0 \]

Rearranging gives:

\[ k^2 - 1 = 0 \]

### Step 7: Solve for \( k \)
Factoring gives:

\[ (k - 1)(k + 1) = 0 \]

Thus, the solutions are:

\[ k = 1 \quad \text{or} \quad k = -1 \]

### Conclusion
The values of \( k \) for which the lines are coplanar are \( k = 1 \) and \( k = -1 \).

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