jmc

Since the axis of symmetry is vertical and the vertex is $(3,2)$, the parabola may also be written as $$y=a(x-3{)}^2+2$$for some value of $a$. Plugging the point $(1,0)$ into this expression gives $$0=a(1-3{)}^2+2=4a+2.$$This tells us $a=-\frac{1}{2}$.

Our equation is $$y=-\frac{1}{2}(x-3{)}^2+2.$$Putting it into $y=a{x}^2+bx+c$ form requires expanding the square, so we get $$y=-\frac{1}{2}(x^2-6x+9)+2=-\frac{1}{2}{x}^2+3x-\frac{5}{2}.$$Our answer is $(a,b,c)=\boxed{\left(-\frac{1}{2},3,-\frac{5}{2}\right)}.$