There can be two cases :

Case I: When 1 row contains 3 balls and rest two contains 1 ball each.

Now, the row which contains 3 balls can be selected out of the 3 rows in ³C₁ ways and in this row number of ways of arrangement = ³C₃. In other two rows, number of ways of arrangement in each = ³C₁.

Thus, number of ways for case I = ³C₁ × ³C_{3 ×} ³C₁ × ³C₁ = 3 × 1 × 3 × 3 = 27

Case II : When 1 row contains 1 ball and rest two rows contain 2 balls each.

This row, containing 1 ball can be selected in 3C, ways and number of ways of arrangement in this row = ³C₁. In other two rows, containing 2 balls each, number of ways of arrangement in each = ³C₂.

Thus, number of ways for case II = ³C₁ × ³C₁ × ³C₂ × ³C₂ = 3 × 3 × 3 × 3 = 81.

As, either of these two cases are possible, hence total number of ways = case I or case II = 27 +81 = 108.

Hence option D is the answer