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 (A) exactly two values (B) exactly thre values (C) any value (D) exactly one value

Options

A. any value
B. exactly one value
C. exactly two value
D. exactly tree value

Answer

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

### Step 1: Write the equations of the lines in parametric form
The first line is given by:
\[ \frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{-k} \]
Let \( t \) be the parameter. Then, we can express the coordinates as:
\[ x = 2 + t, \quad y = 3 + t, \quad z = 4 - kt \]

The second line is given by:
\[ \frac{x-1}{k} = \frac{y-4}{2} = \frac{z-5}{1} \]
Let \( s \) be the parameter. Then, we can express the coordinates as:
\[ x = 1 + ks, \quad y = 4 + 2s, \quad z = 5 + s \]

### Step 2: Identify the direction ratios and points
From the equations, we can identify:
- For the first line, the direction ratios are \( (1, 1, -k) \) and a point on the line is \( (2, 3, 4) \).
- For the second line, the direction ratios are \( (k, 2, 1) \) and a point on the line is \( (1, 4, 5) \).

### Step 3: Set up the determinant for coplanarity
The lines are coplanar if the following determinant is zero:
\[ \begin{vmatrix} 2 - 1 & 3 - 4 & 4 - 5 \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix} = 0 \]

Calculating the first row:
\[ \begin{vmatrix} 1 & -1 & -1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix} \]

### Step 4: Calculate the determinant
Expanding the determinant:
\[ = 1 \cdot \begin{vmatrix} 1 & -k \\ 2 & 1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -k \\ k & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \\ k & 2 \end{vmatrix} \]

Calculating the minors:
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:
\[ = 1(1 + 2k) + (1 + k^2) - (2 - k) = 1 + 2k + 1 + k^2 - 2 + k = k^2 + 3k + 0 \]

### Step 5: Set the determinant to zero
Setting the determinant to zero gives:
\[ k^2 + 3k = 0 \]
Factoring out \( k \):
\[ k(k + 3) = 0 \]
Thus, the solutions are:
\[ k = 0 \quad \text{or} \quad k = -3 \]

### Conclusion
The values of \( k \) for which the lines are coplanar are \( k = 0 \) and \( k = -3 \). Therefore, the answer is that \( k \) can have exactly two values.