If the system fo equations
x+y+z = 5
x + 2y + 3z = 9
`x + 3y + alphaz = beta`
has infinitely many solution, then `beta - alpha` equals

Options

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

Answer

To solve the problem, we need to determine the values of \(\alpha\) and \(\beta\) such that the system of equations has infinitely many solutions. This occurs when the determinant of the coefficients of the system is zero.

### Step-by-Step Solution:

1. **Write the System of Equations:**
The given equations are:
\[ \begin{align*} (1) & \quad x + y + z = 5 \\ (2) & \quad x + 2y + 3z = 9 \\ (3) & \quad x + 3y + \alpha z = \beta \end{align*} \]

2. **Form the Coefficient Matrix:**
The coefficient matrix \(A\) for the system is:
\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha \end{bmatrix} \]

3. **Calculate the Determinant of the Coefficient Matrix:**
We need to find the determinant \(D\) of the matrix \(A\):
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha \end{vmatrix} \]
Expanding this determinant:
\[ D = 1 \cdot \begin{vmatrix} 2 & 3 \\ 3 & \alpha \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 3 \\ 1 & \alpha \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
\[ \begin{vmatrix} 2 & 3 \\ 3 & \alpha \end{vmatrix} = 2\alpha - 9 \]
\[ \begin{vmatrix} 1 & 3 \\ 1 & \alpha \end{vmatrix} = \alpha - 3 \]
\[ \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = 3 - 2 = 1 \]

Thus, we have:
\[ D = (2\alpha - 9) - (\alpha - 3) + 1 = 2\alpha - 9 - \alpha + 3 + 1 = \alpha - 5 \]

4. **Set the Determinant to Zero for Infinitely Many Solutions:**
For the system to have infinitely many solutions, we set the determinant \(D\) to zero:
\[ \alpha - 5 = 0 \implies \alpha = 5 \]

5. **Substitute \(\alpha\) into the Third Equation:**
Now, we substitute \(\alpha = 5\) into the third equation to find \(\beta\):
\[ x + 3y + 5z = \beta \]

6. **Form the New Coefficient Matrix for \(\beta\):**
The new coefficient matrix with \(\beta\) becomes:
\[ B = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \beta \end{bmatrix} \]
We need to find the determinant \(D_3\):
\[ D_3 = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \beta \end{vmatrix} \]

Expanding this determinant:
\[ D_3 = 1 \cdot \begin{vmatrix} 2 & 3 \\ 3 & \beta \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 3 \\ 1 & \beta \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \]

Calculating these determinants:
\[ \begin{vmatrix} 2 & 3 \\ 3 & \beta \end{vmatrix} = 2\beta - 9 \]
\[ \begin{vmatrix} 1 & 3 \\ 1 & \beta \end{vmatrix} = \beta - 3 \]
\[ \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = 1 \]

Thus, we have:
\[ D_3 = (2\beta - 9) - (\beta - 3) + 1 = 2\beta - 9 - \beta + 3 + 1 = \beta - 5 \]

7. **Set the New Determinant to Zero:**
For the system to remain consistent, we set \(D_3 = 0\):
\[ \beta - 5 = 0 \implies \beta = 13 \]

8. **Calculate \(\beta - \alpha\):**
Finally, we find \(\beta - \alpha\):
\[ \beta - \alpha = 13 - 5 = 8 \]

### Final Answer:
\[ \beta - \alpha = 8 \]