Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 17 / c / ii

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 solutionsThis exists for every real $\beta$.