Suppose that we have an equation $y=ax^2+bx+c$ whose graph is a parabola with vertex $(3,2)$, vertical axis of symmetry, and contains the point $(1,0)$.

What is $(a, b, c)$?
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 $a=-\frac12$.

Our equation is  \[y=-\frac12(x-3)^2+2.\]Putting it into $y=ax^2+bx+c$ form requires expanding the square, so we get  \[y=-\frac12(x^2-6x+9)+2=-\frac12 x^2+3x-\frac52.\]Our answer is $(a, b, c) = \boxed{\left(-\frac{1}{2}, 3, -\frac{5}{2}\right)}.$