Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 11 / b / iii / Solution

The same scalar consistency condition now has $\mathbf{a}\cdot \mathbf{c}=1$, so it reads $0=1$. Hence there is no solution.

Geometrically the image of $I+\mathbf{b}{\mathbf{a}}^T$ is still the plane ${\mathbf{a}}^{\perp }$, while this $\mathbf{c}$ is outside it. Thus the failure is inconsistency, rather than a unique solution with a singular denominator.