= Solution
Each plane rotation satisfies $R_j^TR_j=I$. Hence $U=R_3R_2R_1$ is an <orthogonal matrix> and $D=UA$ is an <upper triangular matrix>. Multiplication by $U^T$ gives the <QR decomposition>
$$
\boxed{A=QD,\qquad Q=R_1^TR_2^TR_3^T,\qquad Q^TQ=I.}
$$
The zero-pair cases already handled make this construction valid for every real $3\times3$ matrix, including rank-deficient matrices. Orthogonality follows from the product identity, not from any invertibility assumption on $A$.
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