\(a^2+b^2+c^2=2\left(a+b+c\right)-3\)
\(\Rightarrow a^2+b^2+c^2-2a-2b-2c+3=0\)
\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Rightarrow a=b=c=1\)
Thay và S ta được:
\(S=1^{2019}+1^{2019}+1^{2019}=1+1+1=3\)