The number of values of $$\alpha$$ for which the system of equations
$$x + y + z = \alpha$$
$$\alpha x + 2\alpha y + 3z = -1$$
$$x + 3\alpha y + 5z = 4$$
is inconsistent, is

The system of equations is:

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

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

$$ x + 3\alpha y + 5z = 4 \quad \cdots (3) $$

Form the coefficient matrix and find its determinant.

$$ D = \begin{vmatrix} 1 & 1 & 1 \\ \alpha & 2\alpha & 3 \\ 1 & 3\alpha & 5 \end{vmatrix} $$

Expanding along the first row:

$$ D = 1(10\alpha - 9\alpha) - 1(5\alpha - 3) + 1(3\alpha^2 - 2\alpha) $$

$$ = \alpha - 5\alpha + 3 + 3\alpha^2 - 2\alpha $$

$$ = 3\alpha^2 - 6\alpha + 3 $$

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

The system may be inconsistent when $$D = 0$$, i.e., $$\alpha = 1$$.

Check if $$\alpha = 1$$ gives an inconsistent system.

With $$\alpha = 1$$:

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

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

$$ x + 3y + 5z = 4 \quad \cdots (3') $$

Subtract (1') from (2'): $$y + 2z = -2 \quad \cdots (4)$$

Subtract (2') from (3'): $$y + 2z = 5 \quad \cdots (5)$$

Equations (4) and (5) are contradictory: $$y + 2z$$ cannot equal both $$-2$$ and $$5$$.

Therefore, the system is inconsistent for $$\alpha = 1$$.

For all other values of $$\alpha$$, $$D \neq 0$$, so the system has a unique solution and is consistent.

Therefore, there is exactly 1 value of $$\alpha$$ for which the system is inconsistent.

The answer is Option B: 1.