The number of balls in the red, blue, yellow bowls are $r, b, y$ respectively. Then $r+b+y=27$ and $15r+3b+18y=1+2+3+\ldots+27=378$. Dividing by 3 and subtracting $r+b+y=27$ gives $4r+5y=99$. Then $r=21-5k$ and $y=3+4k$ for some integer $k$, and so $b=3+k$. But $b\leq 5$ because taking the least 6 integers gives an average value of $3.5$ which is too high. Thus $k=0, 1, 2$ so $(r, b, y)=(21, 3, 3); (16, 4, 7); (11, 5, 11)$. These all work as shown by the sets $r_1=\{2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 20, 21, 22, 23, 24, 25, 26, 27\}$ $b_1=\{1, 3, 5\}$ $y_1=\{17, 18, 19\}$ $r_2=\{3, 6, 7, 8, 9, 10, 11, 12, 13, 14, 22, 23, 24, 25, 26, 27\}$ $b_2=\{1, 2, 4, 5\}$ $y_2=\{15, 16, 17, 18, 19, 20, 21\}$ $r_3=\{6, 7, 8, 9, 10, 11, 12, 24, 25, 26, 27\}$ $b_3=\{1, 2, 3, 4, 5\}$ $y_3=\{13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23\}$