 Solution

Set $s=\mathbf{a}\cdot \mathbf{r}$. Taking the dot product of the vector equation with $\mathbf{a}$ givesHere $\mathbf{a}\cdot \mathbf{b}=5$ and $\mathbf{a}\cdot \mathbf{c}=6$, so $s=1$. Substitution then givesGeometrically the possible vector must lie on the line $\mathbf{c}-s\mathbf{b}$, and the scalar constraint picks one point. Equivalently the rank-one matrix update $I+\mathbf{b}{\mathbf{a}}^T$ is invertible because $1+\mathbf{a}\cdot \mathbf{b}=6\ne 0$. This is the nonsingular case of solvability of an affine rank-one vector equation.