suppose you are right,
since p,q, r are logically symetric,
if you can proof b=3+p,a=2/(6-3p)
then you can also proof b=3+q,a=2/(6-3q)
and b=3+r,a=2/(6-3r)
so you get p=q=r (you know why?) ====> something may be wrong in your calcul!!!
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I do some calculs also, f(5) = 3a(5-b), a root of f(x), you get
1) 3a(5-b) = b, or
2) 3a(5-b) = 2
in both case, you replace a by 2/(6-3p), b by 3+p,you can find p
but you said "p can take any complex number,if one of q,r is not real number"
Absurd! so you are not right
