Let the system of linear equations $$x + 2y + z = 2, \alpha x + 3y - z = \alpha, -\alpha x + y + 2z = -\alpha$$ be inconsistent. Then $$\alpha$$ is equal to

The system of linear equations is:

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

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

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

First, we compute the determinant of the coefficient matrix:

$$D = \begin{vmatrix} 1 & 2 & 1 \\ \alpha & 3 & -1 \\ -\alpha & 1 & 2 \end{vmatrix}$$

Expanding along the first row gives

$$D = 1(3 \times 2 - (-1) \times 1) - 2(\alpha \times 2 - (-1)(-\alpha)) + 1(\alpha \times 1 - 3 \times (-\alpha))$$

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

$$= 7 - 2\alpha + 4\alpha$$

$$= 7 + 2\alpha$$

For the system to be inconsistent, we require $$D = 0$$, which leads to

$$7 + 2\alpha = 0$$

$$\alpha = -\frac{7}{2}$$

Next, with $$\alpha = -\frac{7}{2}$$ we compute the determinant $$D_x$$ obtained by replacing the first column with the constants:

$$D_x = \begin{vmatrix} 2 & 2 & 1 \\ -\frac{7}{2} & 3 & -1 \\ \frac{7}{2} & 1 & 2 \end{vmatrix}$$

Evaluating this determinant gives

$$= 2(6+1) - 2(-7+\frac{7}{2}) + 1(-\frac{7}{2} - \frac{21}{2})$$

$$= 14 - 2(-\frac{7}{2}) + 1(-14)$$

$$= 14 + 7 - 14 = 7 \neq 0$$

Since $$D = 0$$ but $$D_x \neq 0$$, the system is inconsistent (has no solution).

The correct answer is Option D: $$-\dfrac{7}{2}$$.