= Solution
The original PDF has $\mathbf c=(-2,1,1)$ here; the transcription loses its minus sign. Now $\mathbf a\cdot\mathbf b=-1$ and $\mathbf a\cdot\mathbf c=0$, so the scalar condition becomes $0=0$ and leaves $s$ arbitrary. Consequently
$$
\boxed{\mathbf r=\mathbf c-s\mathbf b=(-2-s,1+s,1+s),\qquad s\in\mathbb R}.
$$
The <solution set> is a straight line parallel to $\mathbf b$, equivalently the intersection of $y=z$ and $x+y=-1$.
For the geometric mechanism, let $L=I+\mathbf b\mathbf a^T$. Since $\mathbf a\cdot\mathbf b=-1$, $L^2=L$, $L\mathbf b=0$, and $\mathbf a\cdot L\mathbf r=0$. Thus $L$ is an <oblique projection> onto the <plane> $\mathbf a^\perp$ along $\mathbf b$. The specified $\mathbf c$ lies in this image <plane>, and adding any multiple of the <kernel> direction $\mathbf b$ leaves its image unchanged.
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