If the system of linear equations
$$2x + y - z = 3$$
$$x - y - z = \alpha$$
$$3x + 3y + \beta z = 3$$
has infinitely many solutions, then $$|\alpha + \beta - \alpha\beta|$$ is equal to _________.

We have the system of linear equations:

$$2x + y - z = 3 \quad \cdots (1)$$

$$x - y - z = \alpha \quad \cdots (2)$$

$$3x + 3y + \beta z = 3 \quad \cdots (3)$$

For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero and the system must remain consistent.

Step 1: Set the determinant to zero.

The coefficient matrix is:

$$A = \begin{pmatrix} 2 & 1 & -1 \\ 1 & -1 & -1 \\ 3 & 3 & \beta \end{pmatrix}$$

Expanding the determinant along the first row:

$$\det(A) = 2(-\beta + 3) - 1(\beta + 3) + (-1)(3 + 3)$$

$$= -2\beta + 6 - \beta - 3 - 6 = -3\beta - 3 = -3(\beta + 1)$$

Setting $$\det(A) = 0$$ gives $$\beta = -1$$.

Step 2: Find $$\alpha$$ using the consistency condition.

With $$\beta = -1$$, the system becomes:

$$2x + y - z = 3 \quad \cdots (1)$$

$$x - y - z = \alpha \quad \cdots (2)$$

$$3x + 3y - z = 3 \quad \cdots (3)$$

Note that the left-hand side of equation (3) equals $$2 \times$$ (LHS of eq. 1) $$-$$ (LHS of eq. 2):

$$2(2x + y - z) - (x - y - z) = 4x + 2y - 2z - x + y + z = 3x + 3y - z$$

For consistency, the same linear combination must hold for the right-hand sides:

$$2(3) - \alpha = 3$$

$$6 - \alpha = 3$$

$$\alpha = 3$$

Step 3: Compute the required expression.

With $$\alpha = 3$$ and $$\beta = -1$$:

$$\alpha + \beta - \alpha\beta = 3 + (-1) - (3)(-1) = 3 - 1 + 3 = 5$$

Therefore:

$$|\alpha + \beta - \alpha\beta| = |5| = 5$$