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

Options

A. a) `k=0`
B. b) `k=-1`
C. c) `k=2`
D. d) `k=-3`

Answer

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

### Step 1: Identify Points and Direction Ratios
The first line is given by:
\[ \frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{-k} \]
From this, we can identify a point on the line \( A(2, 3, 4) \) and the direction ratios \( \vec{d_1} = (1, 1, -k) \).

The second line is given by:
\[ \frac{x-1}{k} = \frac{y-4}{2} = \frac{z-5}{1} \]
From this, we can identify a point on the line \( B(1, 4, 5) \) and the direction ratios \( \vec{d_2} = (k, 2, 1) \).

### Step 2: Find the Vector Joining the Two Points
The vector joining points \( A \) and \( B \) is:
\[ \vec{AB} = B - A = (1 - 2, 4 - 3, 5 - 4) = (-1, 1, 1) \]

### Step 3: Set Up the Determinant for Coplanarity
The lines are coplanar if the determinant of the matrix formed by the vectors \( \vec{d_1} \), \( \vec{d_2} \), and \( \vec{AB} \) is zero. The determinant can be set up as follows:
\[ \begin{vmatrix} 1 & 1 & -k \\ k & 2 & 1 \\ -1 & 1 & 1 \end{vmatrix} = 0 \]

### Step 4: Calculate the Determinant
Calculating the determinant, we have:
\[ 1 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} k & 1 \\ -1 & 1 \end{vmatrix} - k \cdot \begin{vmatrix} k & 2 \\ -1 & 1 \end{vmatrix} \]

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

2. \( \begin{vmatrix} k & 1 \\ -1 & 1 \end{vmatrix} = k \cdot 1 - 1 \cdot (-1) = k + 1 \)

3. \( \begin{vmatrix} k & 2 \\ -1 & 1 \end{vmatrix} = k \cdot 1 - 2 \cdot (-1) = k + 2 \)

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

### Step 5: Factor the Equation
Factoring out \( -k \):
\[ -k(k + 3) = 0 \]
Thus, we have:
\[ k = 0 \quad \text{or} \quad k = -3 \]

### Final Answer
The lines are coplanar if:
\[ k = 0 \quad \text{or} \quad k = -3 \]