I presume, #p(x)=ax^2+bx+c, and, p^2(x)=(p(x))^2#.
Now, given that,
#p^2(o)+p^2(1)+p^2(2)+166=6p(o)+12p(1)+22p(2)#.
#:. p^2(0)-6p(0)+p^2(1)-12p(1)+p^(2)-22p(2)=-166#.
On completing squares,
#{p(0)-3}^2+{p(1)-6}^2+{p(2)-11}^2#,
#=9+36+121-166#.
#:. {p(0)-3}^2+{p(1)-6}^2+{p(2)-11}^2=0#.
#:. {p(0)-3}=0, {p(1)-6}=0, &, {p(2)-11}=0#.
#:. p(0)=3, p(1)=6, &, p(2)=11#.
Now, #p(0)=3 rArr c=3...............(star_1)#.
#p(1)=6 rArr a+b+c=6."Then, by "(star_1), a+b=3....(star_2)#.
#"Similarly, from "p(2)=11," we get, "2a+b=4...(star_3)#.
#"Altogether "(star_1), (star_2) and (star_3) rArr (a,b,c)=(1,2,3)#.
Consequently, #a-b+c=2#, showing that, (A) is true.
Finally, to find #p_min#, let us observe that,
#p(x)=x^2+2x+3=(x+1)^2+2#.
#because AA x in RR, :. (x+1)^2 ge 0, &, :., (x+1)^2+2 ge 2#.
Clearly, #p_min=2#, meaning that, (B) is also true.
Enjoy Maths.!
