If plane `x + 4y - 2z = 1, x + 7y – 5z = beta, x + 5y + alphaz = 5` intersects in a line `(R xx R xx R)` then `alpha + beta` is equal to

Options

A. `2`
B. `10`
C. `0`
D. `-10`

Answer

To solve the problem, we need to find the values of \( \alpha \) and \( \beta \) such that the three given planes intersect in a line. This condition is satisfied when the determinant of the coefficients of the variables \( x, y, z \) is equal to zero.

### Step-by-Step Solution:

1. **Write the equations of the planes:**
\[ \begin{align*} \text{Plane 1:} & \quad x + 4y - 2z = 1 \\ \text{Plane 2:} & \quad x + 7y - 5z = \beta \\ \text{Plane 3:} & \quad x + 5y + \alpha z = 5 \end{align*} \]

2. **Set up the determinant of the coefficients:**
The coefficients of \( x, y, z \) from the three planes can be arranged in a matrix:
\[ \begin{vmatrix} 1 & 4 & -2 \\ 1 & 7 & -5 \\ 1 & 5 & \alpha \end{vmatrix} \]

3. **Calculate the determinant:**
The determinant can be calculated using the formula for a 3x3 matrix:
\[ D = 1 \begin{vmatrix} 7 & -5 \\ 5 & \alpha \end{vmatrix} - 4 \begin{vmatrix} 1 & -5 \\ 1 & \alpha \end{vmatrix} - 2 \begin{vmatrix} 1 & 7 \\ 1 & 5 \end{vmatrix} \]

Now calculate each of the 2x2 determinants:
\[ \begin{vmatrix} 7 & -5 \\ 5 & \alpha \end{vmatrix} = 7\alpha + 25 \]
\[ \begin{vmatrix} 1 & -5 \\ 1 & \alpha \end{vmatrix} = \alpha + 5 \]
\[ \begin{vmatrix} 1 & 7 \\ 1 & 5 \end{vmatrix} = -2 \]

Substitute these back into the determinant:
\[ D = 1(7\alpha + 25) - 4(\alpha + 5) - 2(-2) \]
\[ D = 7\alpha + 25 - 4\alpha - 20 + 4 \]
\[ D = (7\alpha - 4\alpha) + (25 - 20 + 4) = 3\alpha + 9 \]

4. **Set the determinant equal to zero:**
Since the planes intersect in a line, we set the determinant to zero:
\[ 3\alpha + 9 = 0 \]
\[ 3\alpha = -9 \quad \Rightarrow \quad \alpha = -3 \]

5. **Now find the value of \( \beta \):**
We replace the third column of the determinant with the constants from the equations:
\[ \begin{vmatrix} 1 & 4 & 1 \\ 1 & 7 & \beta \\ 1 & 5 & 5 \end{vmatrix} \]

Calculate this determinant:
\[ D' = 1 \begin{vmatrix} 7 & \beta \\ 5 & 5 \end{vmatrix} - 4 \begin{vmatrix} 1 & \beta \\ 1 & 5 \end{vmatrix} + 1 \begin{vmatrix} 1 & 7 \\ 1 & 5 \end{vmatrix} \]
\[ = 1(7 \cdot 5 - 5 \cdot \beta) - 4(5 - \beta) + (-2) \]
\[ = 35 - 5\beta - 20 + 4\beta - 2 \]
\[ = -\beta + 13 \]

Set this determinant equal to zero:
\[ -\beta + 13 = 0 \quad \Rightarrow \quad \beta = 13 \]

6. **Calculate \( \alpha + \beta \):**
\[ \alpha + \beta = -3 + 13 = 10 \]

### Final Answer:
\[ \alpha + \beta = 10 \]