by using pythagoras theorem for triangle ABC
$(2a)^2=x^2+((a√3)^2)$ (let x be unknown side of triangle ABC)
then x=a

area of big triangle =(1/2)*a*a√3=$(√3/2)*a^2$
we need to remove the area of unshaded region to get area of shaded region
lets find area of unshaded region
from ΔDGH ∠D=30 and HG=b-c
then DG=√3(b-c) (by applying tan formula)
area of ΔDGH=$(√3(b-c)^2 /2)$------(2)
now area of ▭DEFG = √3(b-c)*c
we can write this as=$(√3/2)*(2bc-2c^2)$ -----(3)

total area of unshaded region=$(√3/2)*(b^2-c^2)$

now total area of shaded region will be eqn(1)-eqn(3)=$(√3/2)*(a^2-b^2+c^2)$

option C is correct