System of Linear Equations Condition for No Solution

To determine the condition for which the given system of linear equations has no solution, we use Cramer's Rule for the system:

$$\begin{aligned} x + 2y + z &= 5 \quad \text{--- (1)} \\ 2x + y + \alpha z &= 5 \quad \text{--- (2)} \\ 8x + 4y + \beta z &= 18 \quad \text{--- (3)} \end{aligned}$$

The determinant of the coefficient matrix, denoted as $\Delta$, is given by:

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

Expanding $\Delta$ along the first row:

$$\Delta = 1 \cdot (1 \cdot \beta - 4\alpha) - 2 \cdot (2\beta - 8\alpha) + 1 \cdot (2 \cdot 4 - 8 \cdot 1)$$ $$\Delta = (\beta - 4\alpha) - 2(2\beta - 8\alpha) + (8 - 8)$$ $$\Delta = \beta - 4\alpha - 4\beta + 16\alpha + 0$$ $$\Delta = 12\alpha - 3\beta$$

A system of linear equations has no solution if $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \text{ or } \Delta_z$ is non-zero.

Setting $\Delta = 0$:

$$12\alpha - 3\beta = 0 \implies 3\beta = 12\alpha \implies \frac{\beta}{\alpha} = 4$$

Now, we check $\Delta_z$ to ensure that at least one determinant is non-zero:

$$\Delta_z = \begin{vmatrix} 1 & 2 & 5 \\ 2 & 1 & 5 \\ 8 & 4 & 18 \end{vmatrix}$$

Expanding along the first row:

$$\Delta_z = 1 \cdot (1 \cdot 18 - 4 \cdot 5) - 2 \cdot (2 \cdot 18 - 8 \cdot 5) + 5 \cdot (2 \cdot 4 - 8 \cdot 1)$$ $$\Delta_z = 1(18 - 20) - 2(36 - 40) + 5(8 - 8)$$ $$\Delta_z = -2 - 2(-4) + 0 = -2 + 8 = 6 \neq 0$$

Since $\Delta = 0$ and $\Delta_z \neq 0$, the given system of equations has no solution when $\frac{\beta}{\alpha} = 4$.

Thus, the correct option is B.