Since the plants are indistinguishable, we must only count the number of plants on each window sill. If all the plants are on one window sill, there are $ 3 $ ways to choose which window sill they are on. If $ 5 $ plants are on one window sill and the last is on another, there are $ 3!=6 $ ways to choose which plants go on which window sill. If $ 4 $ plants are on one window sill and the last two are on another, there are $ 3!=6 $ ways to choose which window sill they are on. If $ 4 $ plants are on one window sill and the last two are each on one of the other windows, there are $ 3 $ ways to choose which window the $ 4 $ plants are on. If $ 3 $ plants are on one window and the other $ 3 $ plants are all on another window, there are $ 3 $ ways to choose which window has no plants. If $ 3 $ plants are on one window, $ 2 $ plants on another window, and $ 1 $ plant on the last window, there are $ 3!=6 $ ways to choose which plants are on which windows. If $ 2 $ plants are on each window, there is only one way to arrange them. In total, there are $ 3+6+6+3+3+6+1=\boxed{28} $ ways to arrange the plants on the window sills. See if you can find a faster way to do this problem by considering lining up the plants, and placing two dividers among the plants to separate them into three groups corresponding to the sills.