For the two positive numbers $$a, b$$, if $$a, b$$ and $$\frac{1}{18}$$ are in a geometric progression, while $$\frac{1}{a}$$, 10 and $$\frac{1}{b}$$ are in an arithmetic progression, then, $$16a + 12b$$ is equal to _____.

We are given that $$a, b, \frac{1}{18}$$ are in a geometric progression and that $$\frac{1}{a}, 10, \frac{1}{b}$$ are in an arithmetic progression, and we need to find $$16a + 12b$$.

Because three terms in geometric progression satisfy that the square of the middle term equals the product of the outer terms, we have

$$b^2 = a \times \frac{1}{18} = \frac{a}{18}$$

which implies $$a = 18b^2$$.

Similarly, three terms in arithmetic progression satisfy that twice the middle term equals the sum of the outer terms, so

$$2 \times 10 = \frac{1}{a} + \frac{1}{b}$$

and therefore $$\frac{1}{a} + \frac{1}{b} = 20$$.

Substituting $$a = 18b^2$$ into this equation leads to $$\frac{1}{18b^2} + \frac{1}{b} = 20$$. Multiplying both sides by $$18b^2$$ gives

$$1 + 18b = 360b^2$$

or equivalently

$$360b^2 - 18b - 1 = 0$$

Applying the quadratic formula yields

$$b = \frac{18 \pm \sqrt{324 + 1440}}{720} = \frac{18 \pm \sqrt{1764}}{720} = \frac{18 \pm 42}{720}$$

Since $$a,b > 0$$, we select the positive root $$b = \frac{60}{720} = \frac{1}{12}$$.

Substituting $$b = \frac{1}{12}$$ into $$a = 18b^2$$ gives $$a = 18 \times \frac{1}{144} = \frac{1}{8}$$.

Finally, calculating the required expression, we find

$$16a + 12b = 16 \times \frac{1}{8} + 12 \times \frac{1}{12} = 2 + 1 = 3$$

Hence, the answer is $$3$$.