= Solution
Set $s=\mathbf a\cdot\mathbf r$. Taking the <dot product> of the vector equation with $\mathbf a$ gives
$$
(1+\mathbf a\cdot\mathbf b)s=\mathbf a\cdot\mathbf c.
$$
Here $\mathbf a\cdot\mathbf b=5$ and $\mathbf a\cdot\mathbf c=6$, so $s=1$. Substitution then gives
$$
\boxed{\mathbf r=\mathbf c-\mathbf b=(1,0,0)}.
$$
Geometrically 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 $1+\mathbf a\cdot\mathbf b=6\ne0$. This is the nonsingular case of <solvability of an affine rank-one vector equation>.
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