Step 1: Work out each branch's key value first, before computing every column.
The final operation is a natural join of two intermediate relations on their common columns P, Q, S. A natural join can only produce rows where the P values actually agree, so it helps to track just the P values on each side before doing the full column-by-column join.
Step 2: Track the P value on the left branch.
\(X \Join Y\) only keeps rows where P and Q match between X and Y. Checking each X row against Y gives two survivors: (P1,Q1,R1,2) and (P3,Q3,R2,1).
Applying \(\sigma_{(Q=Q3 \vee R=R2)}\): the P1 row has Q=Q1 and R=R1, neither condition holds, so it is removed. The P3 row has Q=Q3, so it survives.
So the left branch collapses to a single row with P=P3.
Step 3: Track the P value on the right branch.
\(Y \Join Z\) matches on P alone. P1 appears twice in Y and once in Z, giving two joined rows with P=P1: (P1,Q1,2,T1) and (P1,Q2,5,T1). P2 has no partner in Z. P3 gives (P3,Q3,1,T2).
Applying \(\sigma_{(S>1)}\): the two P1 rows have S=2 and S=5, both greater than 1, so both survive. The P3 row has S=1, which fails \(S>1\), so it is removed.
So the right branch collapses to two rows, and both of them have P=P1 (the P3 row was filtered out by the S>1 condition).
Step 4: Compare the P values across the two branches.
The left branch only has P=P3. The right branch only has P=P1. Since a natural join requires equal P values to produce any tuple, and P3 never equals P1, there is no way for any pair of rows to combine.
Step 5: Conclude without needing to check Q or S at all.
Because the P values already disagree completely between the two sides, the join is empty regardless of what the Q and S columns contain, and projecting an empty relation onto P, R, S still gives zero rows. This matches option (D).\[ \boxed{\text{Zero rows}} \]