Step 1: Set up the equal-product condition.
Let the common value of all three products be N, so:
\[ P \times Q \times R = X \times Y \times Z = Q \times A \times Y = N \]
From the third product, \(N = Q \times A \times Y\). Substituting this into the other two:
\[ P \times R = \frac{N}{Q} = A \times Y, \qquad X \times Z = \frac{N}{Y} = A \times Q \]
Step 2: Search for a working set of seven distinct digits.
Try \(A = 2\). Then \(P \times R = 2Y\) and \(X \times Z = 2Q\). Testing \(Q = 9\) and \(Y = 4\), both different from A and from each other:
\(P \times R = 2 \times 4 = 8\), which splits as \(\{P, R\} = \{1, 8\}\), distinct digits, neither equal to 2, 9 or 4.
\(X \times Z = 2 \times 9 = 18\), which splits as \(\{X, Z\} = \{3, 6\}\), again distinct, and not clashing with any digit used so far.
Step 3: Verify all three products actually match.
\(P \times Q \times R = 1 \times 9 \times 8 = 72\).
\(X \times Y \times Z = 3 \times 4 \times 6 = 72\).
\(Q \times A \times Y = 9 \times 2 \times 4 = 72\).
All three equal 72, and the seven digits used, 1, 2, 3, 4, 6, 8, 9, are indeed all different from each other. This is a valid, consistent assignment.
Step 4: Rule out the other options.
If A were 0, every one of the three products would collapse to 0, which would force at least one of P, R (or X, Z) to also be 0 to keep the other products at 0 too, but 0 can only be used once among seven distinct digits, and it cannot sit in both non-overlapping triples at the same time, so A = 0 does not give a consistent set. Trying A = 3 or A = 6 in the same way as Step 2 fails to produce a valid set of seven distinct non-zero digits whose three products all match.
Final Answer:
\[ \boxed{A = 2} \]