If `2x-y-z=3 , x+y-2z=alpha , 3x+3y-betaz=3` has infinite many solution then find the value of `alpha+beta-alphabeta`

Answer

To solve the problem, we need to analyze the given system of equations:

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

For the system to have infinitely many solutions, the determinant of the coefficients of the variables must be zero. We can express this in terms of a determinant \(D\) formed by the coefficients of \(x\), \(y\), and \(z\).

### Step 1: Form the Coefficient Matrix and Calculate the Determinant

The coefficient matrix \(A\) is given by:

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

We calculate the determinant \(D\):

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

Using the determinant formula for a 3x3 matrix:

\[ D = 2 \begin{vmatrix} 1 & -2 \\ 3 & -\beta \end{vmatrix} - (-1) \begin{vmatrix} 1 & -2 \\ 3 & -\beta \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 3 & 3 \end{vmatrix} \]

Calculating the minors:

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

2. \(\begin{vmatrix} 1 & 1 \\ 3 & 3 \end{vmatrix} = 1 \cdot 3 - 1 \cdot 3 = 0\)

Thus, we have:

\[ D = 2(-\beta + 6) + (-1)(-\beta + 6) - 0 \]

Simplifying:

\[ D = -2\beta + 12 + \beta - 6 = -\beta + 6 \]

Setting the determinant to zero for infinite solutions:

\[ -\beta + 6 = 0 \implies \beta = 6 \]

### Step 2: Calculate \(D_1\) for \(\alpha\)

Next, we need to find \(D_1\) by replacing the first column of \(A\) with the constants from the equations:

\[ D_1 = \begin{vmatrix} 3 & -1 & -1 \\ \alpha & 1 & -2 \\ 3 & 3 & -6 \end{vmatrix} \]

Calculating \(D_1\):

\[ D_1 = 3 \begin{vmatrix} 1 & -2 \\ 3 & -6 \end{vmatrix} - (-1) \begin{vmatrix} \alpha & -2 \\ 3 & -6 \end{vmatrix} - 1 \begin{vmatrix} \alpha & 1 \\ 3 & 3 \end{vmatrix} \]

Calculating the minors:

1. \(\begin{vmatrix} 1 & -2 \\ 3 & -6 \end{vmatrix} = 1 \cdot (-6) - (-2) \cdot 3 = -6 + 6 = 0\)

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

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

Thus:

\[ D_1 = 3(0) + (6\alpha - 6) - (3\alpha - 3) \]

Simplifying:

\[ D_1 = 6\alpha - 6 - 3\alpha + 3 = 3\alpha - 3 \]

Setting \(D_1 = 0\):

\[ 3\alpha - 3 = 0 \implies \alpha = 1 \]

### Step 3: Calculate \(\alpha + \beta - \alpha\beta\)

Now we have \(\alpha = 1\) and \(\beta = 6\):

\[ \alpha + \beta - \alpha\beta = 1 + 6 - (1 \cdot 6) = 7 - 6 = 1 \]

### Final Answer

Thus, the value of \(\alpha + \beta - \alpha\beta\) is:

\[ \boxed{1} \]