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 \[ax^2+bx+c=(x+1)^2-1.\] In order to compute $ 100a+10b+c $ we can substitute $ x=10 $ and that gives \[100a+10b+c=(10+1)^2-1=\boxed{120}.\]