jmc

Since the points $(-3,3)$ and $(1,3)$ have the same $y$-value, the axis of symmetry of the parabola must be between these 2 points. The $x$-value halfway between $-3$ and $1$ is $x=-1$. Therefore the vertex of the parabola is equal to $(-1,k)$ for some $k$ and the parabola may also be written as $$a(x+1{)}^2+k.$$ Now we substitute. The point $(1,3)$ gives $$3=a(1+1{)}^2+k,$$ or $$4a+k=3.$$ The point $(0,0)$ gives $$0=a(0+1{)}^2+k$$ or $$a+k=0.$$ Subtracting the second equation from the first gives $$(4a+k)-(a+k)=3-0$$ so $3a=3$, giving $a=1$.

Since $a=1$ and $a+k=0$ we know that $k=-1$ and our parabola is $$a{x}^2+bx+c=(x+1{)}^2-1.$$ In order to compute $100a+10b+c$ we can substitute $x=10$ and that gives $$100a+10b+c=(10+1{)}^2-1=\boxed{120}.$$