 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 conditionsIf $d=0$, then $c=1$ and $a=\pm 1$; if $a=0$, then $d=\pm 1$ and $c=0$ or $2$. In both cases $b=\pm 1$. The eight parameter triples are thereforeThese 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.