Solution
= Solution
When $\alpha=3$, the third <linear equation> is $-3$ times the second, so it adds no constraint. Take $z=t$ freely. The remaining <linear equations> are $x-y=-1-3t$ and $x+6y=1-\beta t$. Solving gives the <straight line> of solutions
$$
\boxed{(x,y,z)=\left(\frac{-5-(\beta+18)t}{7},\ \frac{2+(3-\beta)t}{7},\ t\right),\qquad t\in\mathbb R}.
$$
This exists for every real $\beta$.