 Solution

Use the preceding coefficient identities with the three prescribed plane equations. They immediately giveFor example, its dot product with $\mathbf{a}$ is $s$, because ${\mathbf{a}}^{\prime }\cdot \mathbf{a}=1$ while the other two terms give zero; the other equations follow cyclically. This verifies existence. If two points satisfied all three equations, their difference would be perpendicular to all three members of a basis and therefore zero. Hence the intersection is unique, not merely one possible intersection point.