Let α, β ∈ R be such that the system of linear equations x+2 y+z=5; 2 x+y+α z=5; 8 x+4 y+β z=18 has no solution. Then (β)/(α) is equal to:

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