Let \( Q \) be the foot of perpendicular drawn from the point \( P(1,2,3) \) to the plane \( x + 2y + z = 14 \). If \( R \) is a point on the plane such that \( \angle PRQ = 60^\circ \), then the area of \( \triangle PQR \) is equal to:
\( \dfrac{\sqrt{3}}{2} \)
\( \sqrt{3} \)
\( 2\sqrt{3} \)
\( 3 \)