The set of all non-zero real values of k, for which the lines `(x-4)/(2)=(y-6)/(2)=(z-8)/(-2k^(2))and(x-2)/(2k^(2))=(y-8)/(4)=(z-10)/(2)` are coplanar:

Options

A. is an empty set
B. is a singleton
C. contains two points
D. contains more than two points

Answer

To determine the set of all non-zero real values of \( k \) for which the lines

\[ \frac{x-4}{2} = \frac{y-6}{2} = \frac{z-8}{-2k^2} \]

and

\[ \frac{x-2}{2k^2} = \frac{y-8}{4} = \frac{z-10}{2} \]

are coplanar, we will follow these steps:

### Step 1: Identify Points and Direction Ratios

The first line can be expressed in parametric form. The point on the first line is \( P_1(4, 6, 8) \) and the direction ratios are \( (2, 2, -2k^2) \).

The second line also can be expressed in parametric form. The point on the second line is \( P_2(2, 8, 10) \) and the direction ratios are \( (2k^2, 4, 2) \).

### Step 2: Form Vectors

Let \( \vec{A} \) be the vector from \( P_1 \) to \( P_2 \):
\[ \vec{A} = P_2 - P_1 = (2 - 4, 8 - 6, 10 - 8) = (-2, 2, 2) \]

Let \( \vec{B} \) be the direction vector of the first line:
\[ \vec{B} = (2, 2, -2k^2) \]

Let \( \vec{C} \) be the direction vector of the second line:
\[ \vec{C} = (2k^2, 4, 2) \]

### Step 3: Use the Condition for Coplanarity

The lines are coplanar if the scalar triple product \( \vec{A} \cdot (\vec{B} \times \vec{C}) = 0 \).

### Step 4: Compute the Cross Product \( \vec{B} \times \vec{C} \)

The cross product \( \vec{B} \times \vec{C} \) can be computed using the determinant of a matrix:

\[ \vec{B} \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & -2k^2 \\ 2k^2 & 4 & 2 \end{vmatrix} \]

Calculating this determinant:

\[ = \hat{i} \begin{vmatrix} 2 & -2k^2 \\ 4 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -2k^2 \\ 2k^2 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 2 \\ 2k^2 & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 2 & -2k^2 \\ 4 & 2 \end{vmatrix} = (2)(2) - (-2k^2)(4) = 4 + 8k^2 = 4 + 8k^2 \)
2. \( \begin{vmatrix} 2 & -2k^2 \\ 2k^2 & 2 \end{vmatrix} = (2)(2) - (-2k^2)(2k^2) = 4 + 4k^4 = 4 + 4k^4 \)
3. \( \begin{vmatrix} 2 & 2 \\ 2k^2 & 4 \end{vmatrix} = (2)(4) - (2)(2k^2) = 8 - 4k^2 = 8 - 4k^2 \)

Putting it all together:

\[ \vec{B} \times \vec{C} = \hat{i}(4 + 8k^2) - \hat{j}(4 + 4k^4) + \hat{k}(8 - 4k^2) \]

### Step 5: Compute the Dot Product \( \vec{A} \cdot (\vec{B} \times \vec{C}) \)

Now compute \( \vec{A} \cdot (\vec{B} \times \vec{C}) \):

\[ \vec{A} \cdot (\vec{B} \times \vec{C}) = (-2)(4 + 8k^2) + (2)(-(4 + 4k^4)) + (2)(8 - 4k^2) \]

Expanding this:

\[ = -8 - 16k^2 - 8 - 8k^4 + 16 - 8k^2 \]
\[ = 0 - 24k^2 - 8k^4 \]

Setting this equal to zero for coplanarity:

\[ -8k^4 - 24k^2 = 0 \]

### Step 6: Factor and Solve

Factoring out \( -8k^2 \):

\[ -8k^2(k^2 + 3) = 0 \]

This gives us \( k^2 = 0 \) or \( k^2 + 3 = 0 \). Since \( k^2 = 0 \) gives \( k = 0 \) (not allowed since we want non-zero values), we check \( k^2 + 3 = 0 \):

This has no real solutions since \( k^2 = -3 \) is not possible for real \( k \).

### Conclusion

Thus, there are no non-zero real values of \( k \) for which the lines are coplanar.