To determine the values of \( \alpha \) for which the lines \( L_1 \) and \( L_2 \) are coplanar, we can follow these steps:
### Step 1: Write the equations of the lines in parametric form.
The lines are given as:
- \( L_1: x = 5, \frac{y}{3 - \alpha} = \frac{z}{-2} \)
- \( L_2: x = \alpha, \frac{y}{-1} = \frac{z}{2 - \alpha} \)
We can express these lines in parametric form:
- For \( L_1 \):
- Let \( z = t \) (parameter)
- Then \( y = (3 - \alpha) \cdot \frac{t}{-2} = -\frac{(3 - \alpha)t}{2} \)
- Thus, the parametric equations for \( L_1 \) are:
\[
L_1: (5, -\frac{(3 - \alpha)t}{2}, t)
\]
- For \( L_2 \):
- Let \( z = s \) (parameter)
- Then \( y = -s \)
- Thus, the parametric equations for \( L_2 \) are:
\[
L_2: (\alpha, -s, (2 - \alpha) \cdot \frac{s}{2 - \alpha} = s)
\]
### Step 2: Identify the direction ratios and points.
From the parametric forms, we can identify:
- Direction ratios for \( L_1 \): \( (0, 3 - \alpha, -2) \)
- Direction ratios for \( L_2 \): \( (0, -1, 2 - \alpha) \)
The points on the lines are:
- Point on \( L_1 \): \( (5, 0, 0) \)
- Point on \( L_2 \): \( (\alpha, 0, 0) \)
### Step 3: Set up the determinant for coplanarity.
The lines \( L_1 \) and \( L_2 \) are coplanar if the scalar triple product of the direction ratios and the vector connecting the two points is zero. This can be expressed as a determinant:
\[
\begin{vmatrix}
5 - \alpha & 0 & 0 \\
0 & 3 - \alpha & -2 \\
0 & -1 & 2 - \alpha
\end{vmatrix} = 0
\]
### Step 4: Calculate the determinant.
The determinant simplifies to:
\[
(5 - \alpha) \begin{vmatrix}
3 - \alpha & -2 \\
-1 & 2 - \alpha
\end{vmatrix}
\]
Calculating the 2x2 determinant:
\[
= (3 - \alpha)(2 - \alpha) - (-2)(-1) = (3 - \alpha)(2 - \alpha) - 2
\]
Expanding this:
\[
= 6 - 3\alpha - 2\alpha + \alpha^2 - 2 = \alpha^2 - 5\alpha + 4
\]
Thus, the determinant becomes:
\[
(5 - \alpha)(\alpha^2 - 5\alpha + 4) = 0
\]
### Step 5: Solve for \( \alpha \).
Setting the factors to zero gives:
1. \( 5 - \alpha = 0 \) → \( \alpha = 5 \)
2. \( \alpha^2 - 5\alpha + 4 = 0 \)
Using the quadratic formula:
\[
\alpha = \frac{5 \pm \sqrt{(5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm 3}{2}
\]
This gives:
\[
\alpha = \frac{8}{2} = 4 \quad \text{and} \quad \alpha = \frac{2}{2} = 1
\]
### Final Values of \( \alpha \):
The values of \( \alpha \) for which the lines are coplanar are \( \alpha = 1, 4, 5 \).
### Answer:
The correct options are:
- a. \( 1 \)
- d. \( 4 \)
