\(a^2+b^2+c^2+14=2a+4b+6c\\ < =>a^2-2a+1+b^2-4b+4+c^2-6c+9=0\\ < =>\left(a-1\right)^2+\left(b-2\right)^2+\left(c-3\right)^2=0\\ < =>\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-2\right)^2=0\\\left(c-3\right)^2=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}a=1\\b=2\\c=3\end{matrix}\right.\)
vậy a + b + c = 1 + 2 + 3 = 6

