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 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 coplanar, we can follow these steps:

### Step 1: Identify Points and Direction Vectors

From the first line, we can identify a point and a direction vector:
- Point \( P_1 = (1, 2, 3) \)
- Direction vector \( \vec{d_1} = (2, 3, 4) \)

From the second line, we can identify another point and a direction vector:
- Point \( P_2 = (3, 5, 7) \)
- Direction vector \( \vec{d_2} = (2, \alpha, \alpha + 2) \)

### Step 2: Find the Vector Between the Points

Next, we find the vector between the two points \( P_1 \) and \( P_2 \):
\[ \vec{P_1P_2} = P_2 - P_1 = (3 - 1, 5 - 2, 7 - 3) = (2, 3, 4) \]

### Step 3: Set Up the Determinant for Coplanarity

For the lines to be coplanar, the scalar triple product must be zero. This can be represented using the determinant of a matrix formed by the direction vectors and the vector between the points:

\[ \begin{vmatrix} 2 & 3 & 4 \\ 2 & \alpha & \alpha + 2 \\ 2 & 3 & 4 \end{vmatrix} = 0 \]

### Step 4: Calculate the Determinant

Calculating the determinant, we can simplify it:

1. Notice that the first and third rows are identical:
\[ \begin{vmatrix} 2 & 3 & 4 \\ 2 & \alpha & \alpha + 2 \\ 2 & 3 & 4 \end{vmatrix} = 0 \]

Since two rows are identical, the determinant is always zero, regardless of the value of \( \alpha \).

### Step 5: Conclusion

Since the determinant is always zero, the lines are coplanar for all values of \( \alpha \). Therefore, the complete set of values of \( \alpha \) is:

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