The lines `(x-2)/1=(y-3)/1=(z-4)/(-k) and (x-1)/k=(y-4)/2=(z-5)/1` are coplanar if a. `k=1or-1` b. `k=0or-3` c. `k=3or-3` d. `k=0or-1`

Options

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

Explanation

To determine the values of \( k \) for which the given lines are coplanar, we will use the condition that the determinant formed by the direction ratios and the coordinates of points on the lines must equal zero.

### Step 1: Identify the lines and their parameters
The first line is given by:
\[ \frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{-k} \]
This can be represented in parametric form as:
- Point on the first line: \( (2, 3, 4) \)
- Direction ratios of the first line: \( (1, 1, -k) \)

The second line is given by:
\[ \frac{x-1}{k} = \frac{y-4}{2} = \frac{z-5}{1} \]
This can be represented in parametric form as:
- Point on the second line: \( (1, 4, 5) \)
- Direction ratios of the second line: \( (k, 2, 1) \)

### Step 2: Set up the determinant for coplanarity
The lines are coplanar if the following determinant is equal to zero:
\[ \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0 \]
Substituting the values:
- \( x_2 - x_1 = 1 - 2 = -1 \)
- \( y_2 - y_1 = 4 - 3 = 1 \)
- \( z_2 - z_1 = 5 - 4 = 1 \)

The determinant becomes:
\[ \begin{vmatrix} -1 & 1 & 1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant:
\[ D = -1 \begin{vmatrix} 1 & -k \\ 2 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & -k \\ k & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \\ k & 2 \end{vmatrix} \]

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

Substituting these back into the determinant:
\[ D = -1(1 + 2k) - 1(1 + k^2) + (2 - k) \]
\[ D = -1 - 2k - 1 - k^2 + 2 - k \]
\[ D = -k^2 - 3k = 0 \]

### Step 4: Solve the equation
Factoring the equation:
\[ -k(k + 3) = 0 \]
This gives us:
\[ k = 0 \quad \text{or} \quad k + 3 = 0 \Rightarrow k = -3 \]

### Conclusion
The values of \( k \) for which the lines are coplanar are \( k = 0 \) or \( k = -3 \).

Thus, the correct option is:
**b. \( k = 0 \) or \( k = -3 \)**.