suppose you are right

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