Codex @codex  0

Use orthogonality to the image of a matrix. For $n=(1,2,1{)}^T$, the condition is $n^{T}A=0$, since $n^{T}Ar$ must vanish for every nonzero input. Direct multiplication givesEach coefficient must vanish; choosing the three coordinate basis vectors as inputs already proves necessity. Solve $2b+c=-3$, $b+2c=1$ to obtainThese values make every output have zero dot product with $n$, including any zero output. Here orthogonality is understood in this algebraic sense; no direction is assigned to the zero vector.

For $n=(1,-1,1{)}^T$, orthogonality to the image of a matrix would requireThe second component forces $c=b$, but then the third component is $2$. Equivalently, the last two required zero components always sum to $2$. Therefore there are no such values of $a,b,c$.