If the system of equations
`x+ y+z = 5`
`x+ 2y + 3z =9`
`x +3y + az = beta`
has infinitely many solutions, then `beta-alpha` equals

Options

A. `8`
B. `18`
C. `21`
D. `5`

Explanation

To solve the given system of equations and find the value of \( \beta - \alpha \), we will follow these steps:

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \( x + y + z = 5 \)
2. \( x + 2y + 3z = 9 \)
3. \( x + 3y + az = \beta \)

We can represent this system in matrix form \( AX = B \), where:
\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & a \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ 9 \\ \beta \end{bmatrix} \]

### Step 2: Find the determinant of matrix A
To have infinitely many solutions, the determinant of the coefficient matrix \( A \) must be zero. We calculate the determinant \( D \) of matrix \( A \):
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & a \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the formula for the determinant of a 3x3 matrix:
\[ D = 1 \cdot (2a - 9) - 1 \cdot (1a - 3) + 1 \cdot (3 - 2) \]
This simplifies to:
\[ D = 2a - 9 - a + 3 + 1 = a - 5 \]

### Step 4: Set the determinant to zero
For the system to have infinitely many solutions, we set the determinant to zero:
\[ a - 5 = 0 \implies a = 5 \]

### Step 5: Substitute \( a \) back into the third equation
Now that we have \( a = 5 \), we substitute this back into the third equation:
\[ x + 3y + 5z = \beta \]

### Step 6: Find the determinant for the modified system
Now, we need to find the determinant \( D_3 \) for the modified system:
\[ D_3 = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{vmatrix} \]

Calculating \( D_3 \):
\[ D_3 = 1 \cdot (2 \cdot 5 - 3 \cdot 3) - 1 \cdot (1 \cdot 5 - 3 \cdot 1) + 1 \cdot (1 \cdot 3 - 2 \cdot 1) \]
This simplifies to:
\[ D_3 = 1 \cdot (10 - 9) - 1 \cdot (5 - 3) + 1 \cdot (3 - 2) = 1 - 2 + 1 = 0 \]

### Step 7: Set the new determinant to zero
For the modified system to also have solutions, we set the determinant equal to zero:
\[ D_3 = 0 \implies \beta - 13 = 0 \implies \beta = 13 \]

### Step 8: Calculate \( \beta - \alpha \)
Now we have \( \alpha = 5 \) and \( \beta = 13 \):
\[ \beta - \alpha = 13 - 5 = 8 \]

### Final Answer
Thus, the value of \( \beta - \alpha \) is \( \boxed{8} \).

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