Solution (By Examveda Team)
$$\eqalign{
& p{\left( {x + y} \right)^2} = 5{\text{ and }}q{\left( {x - y} \right)^2} = 3 \cr
& {\text{Put the value of }}x = 2{\text{ and }}y = 1 \cr
& p{\left( {2 + 1} \right)^2} = 5{\text{ and }}q{\left( {2 - 1} \right)^2} = 3 \cr
& p = \frac{5}{9},\,q = 3 \cr
& \to {p^2}{\left( {x + y} \right)^2} + 4pqxy - {q^2}{\left( {x - y} \right)^2} \cr
& = {\left( {\frac{5}{9}} \right)^2}{\left( {2 + 1} \right)^2} + 4 \times \frac{5}{9} \times 3 \times 2 \times 1 - {\left( 3 \right)^2}{\left( {2 - 1} \right)^2} \cr
& = \frac{{25}}{{81}} \times 9 + \frac{{40}}{3} - 9 \cr
& = \frac{{25}}{9} + \frac{{40}}{3} - 9 \cr
& = \frac{{25 + 120 - 81}}{9} \cr
& = \frac{{64}}{9} \cr
& {\text{Put the value of }}p{\text{ and }}q{\text{ in option A}} \cr
& {\text{option A }} \to {\text{2}}\left( {p + q} \right) \cr
& = 2\left( {\frac{5}{9} + 3} \right) \cr
& = 2 \times \frac{{32}}{9} \cr
& = \frac{{64}}{9} \cr
& {\text{Option A is satisfied}} \cr
& {\text{So, 2}}\left( {p + q} \right)\,{\text{is answer}} \cr} $$
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