The complete set of values of `alpha` for which the lines `(x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x-3)/(2) =(y-5)/(alpha)=(z-7)/(alpha+2)` are concurrent and coplanar is

Options

A. `{2, 3}`
B. `{0, 3}`
C. `[-2, 3]`
D. R

Answer

To determine the complete set of values of \( \alpha \) for which the lines

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

and

\[ \frac{x-3}{2} = \frac{y-5}{\alpha} = \frac{z-7}{\alpha + 2} \]

are concurrent and coplanar, we will follow these steps:

### Step 1: Write the parametric equations of the lines

For the first line, let \( \lambda \) be the parameter:

- \( x = 1 + 2\lambda \)
- \( y = 2 + 3\lambda \)
- \( z = 3 + 4\lambda \)

For the second line, let \( \mu \) be the parameter:

- \( x = 3 + 2\mu \)
- \( y = 5 + \alpha\mu \)
- \( z = 7 + (\alpha + 2)\mu \)

### Step 2: Set the equations equal to find a common point

Since the lines are concurrent, there exists a common point for some values of \( \lambda \) and \( \mu \). We equate the \( x \), \( y \), and \( z \) coordinates:

1. From \( x \):
\[ 1 + 2\lambda = 3 + 2\mu \quad \Rightarrow \quad 2\lambda - 2\mu = 2 \quad \Rightarrow \quad \lambda - \mu = 1 \quad \text{(Equation 1)} \]

2. From \( y \):
\[ 2 + 3\lambda = 5 + \alpha\mu \quad \Rightarrow \quad 3\lambda - \alpha\mu = 3 \quad \text{(Equation 2)} \]

3. From \( z \):
\[ 3 + 4\lambda = 7 + (\alpha + 2)\mu \quad \Rightarrow \quad 4\lambda - (\alpha + 2)\mu = 4 \quad \text{(Equation 3)} \]

### Step 3: Solve the equations

From Equation 1, we have \( \lambda = \mu + 1 \). Substitute \( \lambda \) in Equations 2 and 3:

- Substitute in Equation 2:
\[ 3(\mu + 1) - \alpha\mu = 3 \quad \Rightarrow \quad 3\mu + 3 - \alpha\mu = 3 \quad \Rightarrow \quad (3 - \alpha)\mu = 0 \]

- This gives us two cases:
1. \( \mu = 0 \)
2. \( 3 - \alpha = 0 \) or \( \alpha = 3 \)

- Substitute in Equation 3:
\[ 4(\mu + 1) - (\alpha + 2)\mu = 4 \quad \Rightarrow \quad 4\mu + 4 - \alpha\mu - 2\mu = 4 \quad \Rightarrow \quad (4 - \alpha - 2)\mu = 0 \]

### Step 4: Analyze the cases

From both cases, we conclude:

1. If \( \mu = 0 \), the lines are concurrent for any \( \alpha \).
2. If \( \alpha = 3 \), the lines are also concurrent.

### Conclusion

Thus, the complete set of values of \( \alpha \) for which the lines are concurrent and coplanar is:

\[ \alpha \in \mathbb{R} \]