The lines `(x-2)/1=(y-3)/2=(z-4)/(-2k)`and
`(x-1)/k=(y-2)/3=(z-6)/1` are coplanar if

Options

A. `k = 0, k = 2`
B. `k=(5+sqrt(15))/2, k =3`
C. `k=(5 pm sqrt(15))/2`
D. `k = 3, k = 5`

Answer

To determine the value of \( k \) such that the lines

\[ \frac{x-2}{1} = \frac{y-3}{2} = \frac{z-4}{-2k} \]

and

\[ \frac{x-1}{k} = \frac{y-2}{3} = \frac{z-6}{1} \]

are coplanar, we can follow these steps:

### Step 1: Identify the parameters of the lines

For the first line, we can express it in the form:

- Point \( P_1 = (2, 3, 4) \)
- Direction ratios \( a_1 = 1, b_1 = 2, c_1 = -2k \)

For the second line, we have:

- Point \( P_2 = (1, 2, 6) \)
- Direction ratios \( a_2 = k, b_2 = 3, c_2 = 1 \)

### Step 2: Use the coplanarity condition

The lines are coplanar if the following determinant is equal to zero:

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

Substituting the values we have:

- \( x_2 - x_1 = 1 - 2 = -1 \)
- \( y_2 - y_1 = 2 - 3 = -1 \)
- \( z_2 - z_1 = 6 - 4 = 2 \)

Thus, the determinant becomes:

\[ \begin{vmatrix} -1 & -1 & 2 \\ 1 & 2 & -2k \\ k & 3 & 1 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant

Calculating the determinant, we have:

\[ = -1 \begin{vmatrix} 2 & -2k \\ 3 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & -2k \\ k & 1 \end{vmatrix} + 2 \begin{vmatrix} 1 & 2 \\ k & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 2 & -2k \\ 3 & 1 \end{vmatrix} = 2 \cdot 1 - (-2k) \cdot 3 = 2 + 6k = 2 + 6k \)

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

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

Putting it all together:

\[ -1(2 + 6k) + 1(1 + 2k^2) + 2(3 - 2k) = 0 \]

### Step 4: Simplify the equation

This simplifies to:

\[ -2 - 6k + 1 + 2k^2 + 6 - 4k = 0 \]

Combining like terms:

\[ 2k^2 - 10k + 5 = 0 \]

### Step 5: Solve the quadratic equation

Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 2, b = -10, c = 5 \):

\[ k = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 2 \cdot 5}}{2 \cdot 2} \]

Calculating the discriminant:

\[ = \frac{10 \pm \sqrt{100 - 40}}{4} = \frac{10 \pm \sqrt{60}}{4} = \frac{10 \pm 2\sqrt{15}}{4} = \frac{5 \pm \sqrt{15}}{2} \]

### Final Result

Thus, the values of \( k \) for which the lines are coplanar are:

\[ k = \frac{5 + \sqrt{15}}{2} \quad \text{and} \quad k = \frac{5 - \sqrt{15}}{2} \]