Solution:

We observe that
$$Q = a^{3} + b^{3} + c^{3} - 3 a b c = \frac{1}{2}(a + b + c)\left((a - b)^{2} + (b - c)^{2} + (c - a)^{2}\right)$$
Since we are looking for the least positive value taken by $Q$, it follows that $a, b, c$ are not all equal. Thus $a + b + c \geq 1 + 1 + 2 = 4$ and $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} \geq 1 + 1 + 0 = 2$. Thus we see that $Q \geq 4$. Taking $a = 1$, $b = 1$ and $c = 2$, we get $Q = 4$. Therefore the least value of $Q$ is $4$ and this is achieved only by $a + b + c = 4$ and $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 2$. The triples for which $Q = 4$ are therefore given by
$$(a, b, c) = (1, 1, 2),\ (1, 2, 1),\ (2, 1, 1)$$