jmc

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.