∴a=k(b+c)

b=k(c+a)

c=k(a+b)

∴a+b+c=2k(a+b+c)

① when a+b+c = 0, a+b=-c, and the original formula = [2 (a+b)+c]/[(a+b)-3c] = (-2c+c)/(-c-3c) =1.

② when a+b+c ≠ 0, k= 1/2, that is, a+b=2c, and the original formula =(4c+c)/(2c-3c)=-5.

7. Solution: ∫a+2b = 2006

∴ Original formula =3(a+2b)? /2(a+2b)= 3(a+2b)/2 = 3×2006/2 = 3009