jmc

Since the axis of symmetry is vertical and the vertex is $(5,3)$, the parabola may also be written as $$y=a(x-5{)}^2+3$$for some value of $a$. Plugging the point $(2,0)$ into this equation gives $$0=a(2-5{)}^2+3=9a+3.$$This tells us $a=-\frac{1}{3}$.

Our equation is $$y=-\frac{1}{3}(x-5{)}^2+3.$$Putting it $y=a{x}^2+bx+c$ form requires expanding the square, so we get $$y=-\frac{1}{3}(x^2-10x+25)+3=-\frac{1}{3}{x}^2+\frac{10}{3}x-\frac{16}{3}.$$Therefore, $a+b+c=\boxed{-\frac{7}{3}}$.