**Explanation:**

Let x, y, z be three distinct positive integers satisfying the given conditions. We may assume that x < y < z. Thus we have three relations:

**Alternative Solution:**

We first observe that (1, a, b) is not a solution whenever 1 < a < b. Otherwise we should have 1/a + 1/b = l/1 = l for some integer. Hence we obtain (a + b)/ab = l showing that a/b and b/a. Thus a = b contradicting a ≠ b .

Thus the least number should be 2. It is easy to verify that (2, 3, 4) and (2, 3, 5) are not solutions and (2, 3, 6) satisfies all the conditions. (We may observe (2, 4, 5) is also not a solution). Since 3 + 4 + 5 = 12 >11 = 2 + 3 + 6, it follows that (2, 3, 6) has the required minimality.