If the lines `(x-2)/(1)=(y-3)/(1)=(z-4)/(-k)` and `(x-1)/(k)=(y-4)/(2)=(z-5)/(1)` are coplanar, then k can have

Options

A. any value
B. exactly one value
C. exactly two values
D. `exactly three values

Answer

To determine the values of \( k \) for which the given lines are coplanar, we can use the condition for coplanarity of two lines in three-dimensional space. The lines are given in symmetric form:

1. Line 1: \(\frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{-k}\)
2. Line 2: \(\frac{x-1}{k} = \frac{y-4}{2} = \frac{z-5}{1}\)

### Step 1: Identify the direction ratios and points on each line

From the first line, we can extract:
- A point on Line 1: \( P_1(2, 3, 4) \)
- Direction ratios of Line 1: \( (1, 1, -k) \)

From the second line, we can extract:
- A point on Line 2: \( P_2(1, 4, 5) \)
- Direction ratios of Line 2: \( (k, 2, 1) \)

### Step 2: Set up the determinant for coplanarity

To check for coplanarity, we can use the determinant of a matrix formed by the vectors \( \overrightarrow{P_1P_2} \) (the vector from point \( P_1 \) to point \( P_2 \)), and the direction ratios of both lines.

The vector \( \overrightarrow{P_1P_2} \) is given by:
\[ \overrightarrow{P_1P_2} = P_2 - P_1 = (1-2, 4-3, 5-4) = (-1, 1, 1) \]

Now, we can set up the determinant:
\[ \begin{vmatrix} -1 & 1 & 1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant

Calculating the determinant:
\[ \begin{vmatrix} -1 & 1 & 1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix} = -1 \begin{vmatrix} 1 & -k \\ 2 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & -k \\ k & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \\ k & 2 \end{vmatrix} \]

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

Putting it all together:
\[ -1(1 + 2k) - (1 + k^2) + (2 - k) = 0 \]
This simplifies to:
\[ -1 - 2k - 1 - k^2 + 2 - k = 0 \]
Combining like terms:
\[ -k^2 - 3k = 0 \]

### Step 4: Factor the equation

Factoring out \( k \):
\[ k(-k - 3) = 0 \]

### Step 5: Solve for \( k \)

Setting each factor to zero gives:
1. \( k = 0 \)
2. \( -k - 3 = 0 \) or \( k = -3 \)

Thus, the values of \( k \) for which the lines are coplanar are:
\[ k = 0 \quad \text{and} \quad k = -3 \]

### Final Answer:
The values of \( k \) for which the lines are coplanar are \( k = 0 \) and \( k = -3 \).

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