I'm assuming that we have the following :
Let A = (0,0)
Since BE = 2, then AB = 3BE = 6...so....AE = 4
And since angle BAC = 60°, the point B is (6 cos 60°, 6 sin 60°) = (3, 3√3)
And point E is ( 4cos 60°, 4 sin 60°) = (2, 2√3)
And since AD = 3, then AC = 3AD = 9
So D is (3, 0)
So.....the area of BCDE = area of triangle ABC - area of triangle AED
Area of triangle ABC = (1/2)DB * AC = (1/2) (3√3) (9) = (13.5)√3 units^2
And area of triangle AED = (1/2) (AD) (2√3) = (1/2)(3)(2√3) = 3√3 units^2
So...the area of quadrilateral BCDE = [ (13.5)√3 - 3√3] units^2 = (10.5√3) units^2