jmc

We can rewrite the equation of the original parabola as $y=f(x)=a(x-h{)}^2+k$ (for some $a$). After reflection of the parabola, the equation becomes $y=g(x)=-a(x-h{)}^2+k$. Notice that $f(x)+g(x)=2k$. Since $f(1)=a+b+c$ and $g(1)=d+e+f$, we have $a+b+c+d+e+f=f(1)+g(1)=\boxed{2k}$.