I recently offered this slightly hand-wavy argument in response to the linked question regarding linear independence. I wasn't confident enough to make it an answer, so I posted it as a tentative comment, and my question now is whether my argument works. The goal is to find an $x$ that makes $$a_1 = (-1, 1, 0, x)\\ a_2 = (2, -3, 1, 2)\\ a_3 = (1, -2, 1, -1)$$
a linearly independent set, the possibility of which we'll interpret to be implied by the problem. My thought process is basically:
1) If we can make a geometric argument for finding such an $x$ with three vectors in $\mathbb{R}^3$ (at least an argument of the sort in the next step), then it should apply to $\mathbb{R}^4$ even though no such geometry is available,
2) In $\mathbb{R}^3$ (i.e., remove the first element from each vector), the fact that a solution is possible means that $a_1$ is not trapped in the plane spanned by the other vectors, thus there can be at most a single value of $x$ where $a_1$ intersects the plane. The slope of the plane must be rational, since it is determined by ratios of components of $a_2$ and $a_3$, but then picking any irrational $x$ would give $a_1$ an irrational slope, so that it couldn't possibly intersect with the plane.
Does this line of reasoning solve the problem?