Consider the following system of equations
`alpha x+2y+z=1`
`2 alpha x+3y+z=1`
`3x+alpha y+2z=beta`
for some `alpha,beta in R`.Then which of the following is NOT correct

Options

A. It has no solution for `alpha=-1` and `beta!=2`
B. It has no solution for `alpha=-1` and for all `beta in R`
C. It has a solution for all `alpha!=-1` and `beta=2`
D. It has no solution for `alpha=3` and forall `beta!=2`

Answer

To solve the given system of equations and determine which statement is NOT correct, we will analyze the equations step by step.

### Given System of Equations:
1. \( \alpha x + 2y + z = 1 \) (Equation 1)
2. \( 2\alpha x + 3y + z = 1 \) (Equation 2)
3. \( 3x + \alpha y + 2z = \beta \) (Equation 3)

### Step 1: Form the Coefficient Matrix
We will form the coefficient matrix \( A \) from the coefficients of \( x, y, z \) in the equations:

\[ A = \begin{bmatrix} \alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & 2 \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix
To find the conditions for the system to have solutions, we will calculate the determinant of matrix \( A \):

\[ D = \begin{vmatrix} \alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & 2 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix, we calculate:

\[ D = \alpha \begin{vmatrix} 3 & 1 \\ \alpha & 2 \end{vmatrix} - 2 \begin{vmatrix} 2\alpha & 1 \\ 3 & 2 \end{vmatrix} + 1 \begin{vmatrix} 2\alpha & 3 \\ 3 & \alpha \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 3 & 1 \\ \alpha & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot \alpha = 6 - \alpha \)

2. \( \begin{vmatrix} 2\alpha & 1 \\ 3 & 2 \end{vmatrix} = 2\alpha \cdot 2 - 1 \cdot 3 = 4\alpha - 3 \)

3. \( \begin{vmatrix} 2\alpha & 3 \\ 3 & \alpha \end{vmatrix} = 2\alpha \cdot \alpha - 3 \cdot 3 = 2\alpha^2 - 9 \)

Now substituting back into the determinant \( D \):

\[ D = \alpha(6 - \alpha) - 2(4\alpha - 3) + (2\alpha^2 - 9) \]

Expanding this:

\[ D = 6\alpha - \alpha^2 - 8\alpha + 6 + 2\alpha^2 - 9 \]

Combining like terms:

\[ D = (2\alpha^2 - \alpha^2) + (6\alpha - 8\alpha) + (6 - 9) = \alpha^2 - 2\alpha - 3 \]

### Step 3: Set the Determinant to Zero
To find when the system has either no solutions or infinite solutions, we set the determinant to zero:

\[ \alpha^2 - 2\alpha - 3 = 0 \]

Factoring the quadratic:

\[ (\alpha - 3)(\alpha + 1) = 0 \]

Thus, \( \alpha = 3 \) or \( \alpha = -1 \).

### Step 4: Analyze the Cases
- If \( \alpha = 3 \), we need to check if the system has a unique solution or infinite solutions.
- If \( \alpha = -1 \), we substitute back into the equations to check for consistency with \( \beta \).

### Step 5: Find the Value of \( \beta \)
Substituting \( \alpha = -1 \) into the equations, we can find the corresponding \( \beta \):

1. \( -x + 2y + z = 1 \)
2. \( -2x + 3y + z = 1 \)
3. \( 3x - y + 2z = \beta \)

From the first two equations, we can solve for \( z \) in terms of \( x \) and \( y \) and substitute into the third equation to find \( \beta \).

### Conclusion
After substituting and simplifying, we find that for \( \alpha = -1 \), \( \beta \) must equal 2 for the system to have infinite solutions.

### Final Answer
The statement that is NOT correct is that the system has no solution for \( \alpha = -1 \) and for all \( \beta \in \mathbb{R} \).