The total number of positive integral solutions $$(x, y, z)$$ such that $$xyz = 24$$ is:

We need to find the total number of positive integral solutions $$(x, y, z)$$ such that $$xyz = 24$$.

First, we find the prime factorization: $$24 = 2^3 \times 3$$. The divisors of 24 are $$1, 2, 3, 4, 6, 8, 12, 24$$.

For each value of $$x$$ that divides 24, we count the number of ordered pairs $$(y, z)$$ such that $$yz = \frac{24}{x}$$. This count equals the number of divisors of $$\frac{24}{x}$$, denoted $$d\!\left(\frac{24}{x}\right)$$.

Computing for each divisor: when $$x = 1$$, $$yz = 24$$ which has $$d(24) = 8$$ solutions; when $$x = 2$$, $$yz = 12$$ which has $$d(12) = 6$$ solutions; when $$x = 3$$, $$yz = 8$$ which has $$d(8) = 4$$ solutions; when $$x = 4$$, $$yz = 6$$ which has $$d(6) = 4$$ solutions; when $$x = 6$$, $$yz = 4$$ which has $$d(4) = 3$$ solutions; when $$x = 8$$, $$yz = 3$$ which has $$d(3) = 2$$ solutions; when $$x = 12$$, $$yz = 2$$ which has $$d(2) = 2$$ solutions; when $$x = 24$$, $$yz = 1$$ which has $$d(1) = 1$$ solution.

The total number of ordered triples is $$8 + 6 + 4 + 4 + 3 + 2 + 2 + 1 = 30$$.

Therefore, the total number of positive integral solutions is $$\boxed{30}$$.