Let $$\alpha, \beta \in \mathbb{R}$$ be such that the system of linear equations
$$x + 2y + z = 5$$
$$2x + y + \alpha z = 5$$
$$8x + 4y + \beta z = 18$$
has no solution. Then $$\frac{\beta}{\alpha}$$ is equal to :

For the system to have no solution, the determinant of the coefficient matrix must be zero, and the system must be inconsistent.

The coefficient matrix determinant is:

$$D = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta \end{vmatrix}$$

Expanding along the first row:

$$D = 1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(8 - 8)$$

$$= \beta - 4\alpha - 4\beta + 16\alpha + 0 = -3\beta + 12\alpha$$

Setting $$D = 0$$: $$-3\beta + 12\alpha = 0$$, which gives $$\beta = 4\alpha$$.

We verify the system is indeed inconsistent. We check whether $$R_3$$ is a linear combination of $$R_1$$ and $$R_2$$. Let $$R_3 = aR_1 + bR_2$$. From the coefficient columns:

$$a + 2b = 8$$, $$2a + b = 4$$

From the first equation, $$a = 8 - 2b$$. Substituting into the second: $$2(8 - 2b) + b = 4$$, so $$16 - 3b = 4$$, giving $$b = 4$$ and $$a = 0$$.

With $$\beta = 4\alpha$$, the third equation coefficients satisfy $$R_3 = 4R_2$$ (since $$8 = 4 \times 2$$, $$4 = 4 \times 1$$, $$4\alpha = 4 \times \alpha$$). However, for the right-hand side: $$4 \times 5 = 20 \neq 18$$.

Since the coefficient rows are dependent but the augmented matrix is inconsistent, the system has no solution.

Therefore, $$\frac{\beta}{\alpha} = \frac{4\alpha}{\alpha} = 4$$.

Hence, the correct answer is Option 2.