(b) 27√3 cm2.
Let G be the centroid of ΔPQR. Then, PG = 6 cm.
Now, \(\frac{PG}{GS}\) = \(\frac{2}{1}\) ⇒ GS = 3 cm
∴PS = PG + GS = 9 cm …(i)
∴ If a is the length of a side of ΔPQR, then ΔPQR being equilateral, PS⊥QR
∴ Altitude PS = \(\frac{\sqrt3}{2}\) a = 9 (From (i))
⇒ a = \(\frac{9\times2}{\sqrt3}\) = 6√3 cm
∴ Area of equilateral ΔPQR = \(\frac{\sqrt3}{4}\) (a)2
= \(\frac{\sqrt3}{4}\) x (6√3)2 cm2 = 27√3 cm2.