Algebra

when a = b = c = 1, a^2 + b^2 + c^2 = 3
when a = b = 2, c = -1, a^2 + b^2 + c^2 = 9

so, not enough information provided

Not enough information is provided since a b c can be many variables if it is just given that their sum is 3.

If a+b+c=3, then a^2+b^2+c^2>=3 where a, b, c are real.

Proof:

By Schwartz inequality,
(ax1+bx1+cx1)^2<=(a^2+b^2+c^2) (1^2+1^2+1^2)
3^2<=(a^2+b^2+c^2) (3)
a^2+b^2+c^2>=3.