∠BAC=90 ,AD⊥BC,BC= 12，

∴ac= 12cosc,ab= 12sinc,bd= 12sin^c，

△ADC is intercepted by BE, and by Menelaus theorem,

AP/PD*DB/BC*CE/EA= 1, P is the midpoint of AD, AE=3,

∴sin^c*( 12cosc-3)/3= 1，

∴( 1-cos^c)(4cosc- 1)= 1，

4cosc- 1-4(cosc)^3+cos^c= 1，

4(cosC)^2-cos^C-4cosC+2=0,①

Let x = cosc ∈ (- 1, 1) and f (x) = 4x 3-x 2-4x+2, then

f'(x)= 12x^2-2x-4= 12(x+ 1/2)(x-2/3)，

- 1/2 & lt； X & lt2/3 ft (x)

The minimum value of f (x) = f (2/3) = 32/27-4/9-8/3+2 = (32-12-72+54)/27 = 2/27 > 0,

F (- 1) = 1, F (1) = 1, ∴ F (x) has no zero, that is, equation ① has no solution. There is no solution to this problem.