Column I contains rational algebraic expressions and Column II contains possible integers.

Answer

`(A)to(p,q,r,s),(B)to(p,q),(C)t(s)`

Explanation

`Ato(p,q,r,s,),Bto(p,q),Cto(s)`
(A) Let `y=f(x)=x^(3)-6x^(2)+9x+lamda`
`f'(x)=3x^(2)-12x+9=0`
`:.x=1,3`
`f''(x)=6x-12`
`f''(1)lt0` and `f''(3)gt0`

Also `f(0)lt0rarrlamdalt0`
`f(1)gt0`
`implies1-6+9+lamdagt0`
`implieslamda gt-4`......ii and `f(3)lt0` ltbr `implies27-54+27+lamda lt 0`
`implieslamda lt 0`
From Eqs i and ii , iii we get
`-4ltladmalt0`
`implies-3lt lamda +1lt1`
`:.[lamda+1]=-3,-2,-1,0`
`:.|[lamda+1]|=3,2,1,0`
(B) `:' x^(2)+x+1gt0, AA x epsilon R`
Given `-3lt(x^(2)-lamdax-2)/(x^(2)+x+1)lt2`
`implies-3x^(2)-3x-3ltx^(2)-lamdax-2lt2x^(2)+2x+2`
`implies4x^(2)-(lamda+2)x+4gt0`
`:.(lamda-3)^(2)-4.4.1lt0`
and `(lamda+2)^(2)-4.1.4lt0`
`implies(lamda-3)^(2)-4^(2)lt0`
and `(lamda+2)^(2)-4^(2)lt0`
`implies-4ltlamda-3lt4`
and `-4ltlamda+2lt4`
or `-1ltlamdalt7`
an `-6ltlamdalt2`
We get `-1ltlamdalt2`
`:.[lamda]=-1,0,1`
`implies|[lamda]|=0,1`
(C)`:'(b-c)+(c-a)+(a-b)=0`
`:.x=1` is a root of
`(b-c)x^(2)_(c-a)x+(a-b)=0`
Also `x=1` satisfies
`x^(2)+lamdax+1=0`
`implies1+lamda+1=0`
`:.lamda=-2`
Now`lamda-1=-3`
`[lamda-1]=-3`
`implies|[lamda-1]|=3`