Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such \(\rm ​\frac{{QA}}{{AR}}=\frac{{RB}}{{BP}}=\frac{{PC}}{{CQ}}=\frac{1}{2}\). Then \(\rm \frac{\operatorname{Area}(\triangle {PQR})}{\operatorname{Area}(\triangle {ABC})}\) is equal to

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