= Solution
Put $d=c-1$. The <matrix> is real and symmetric, so <orthogonality> means $A^{\mathsf T}A=A^2=I$. Multiplication gives the scalar conditions
$$
a^2+d^2=1,\qquad b^2=1,\qquad2ad=0.
$$
If $d=0$, then $c=1$ and $a=\pm1$; if $a=0$, then $d=\pm1$ and $c=0$ or $2$. In both cases $b=\pm1$. The eight parameter triples are therefore
$$
\boxed{\begin{gathered}
(a,b,c)^{\mathsf T}\in\{(1,1,1)^{\mathsf T},(1,-1,1)^{\mathsf T},(-1,1,1)^{\mathsf T},(-1,-1,1)^{\mathsf T},\\
(0,1,0)^{\mathsf T},(0,-1,0)^{\mathsf T},(0,1,2)^{\mathsf T},(0,-1,2)^{\mathsf T}\}.
\end{gathered}}
$$
These conditions are sufficient as well as necessary, since they make every entry of $A^{\mathsf T}A$ equal to the corresponding entry of the <identity matrix>.
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