If the points $P = \hat{i} + 2 \hat{j}$,$Q = 4 \hat{i} + 6 \hat{j}$,$R = 5 \hat{i} + 7 \hat{j}$,and $S = a \hat{i} + b \hat{j}$ are the consecutive vertices of a parallelogram $PQRS$,then:

  • A
    $a = 2, b = 4$
  • B
    $a = 3, b = 4$
  • C
    $a = 2, b = 3$
  • D
    $a = 3, b = 5$

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