MathematicsMedium249×since 2002Q2600Let QQQ be the foot of perpendicular drawn from the point P(1,2,3)P(1,2,3)P(1,2,3) to the plane x+2y+z=14x+2 y+z=14x+2y+z=14. If RRR is a point on the plane such that ∠PRQ=60∘\angle P R Q=60^{\circ}∠PRQ=60∘, then the area of △PQR\triangle P Q R△PQR is equal to :A32\frac{\sqrt{3}}{2}23B3\sqrt{3}3C232 \sqrt{3}23D3Check answerSkip