Let $$0 < z < y < x$$ be three real numbers such that $$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$$ are in an arithmetic progression and $$x, \sqrt{2}y, z$$ are in a geometric progression. If $$xy + yz + zx = \frac{3}{\sqrt{2}}xyz$$, then $$3(x + y + z)^2$$ is equal to _____.

Given: $$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$$ are in AP, so $$\frac{1}{x} + \frac{1}{z} = \frac{2}{y}$$ $$-(1)$$

Also, $$x, \sqrt{2}y, z$$ are in GP, so $$(\sqrt{2}y)^2 = xz$$, which gives $$2y^2 = xz$$ $$-(2)$$

Given condition: $$xy + yz + zx = \frac{3}{\sqrt{2}}xyz$$

Dividing both sides by $$xyz$$:

$$\frac{1}{z} + \frac{1}{x} + \frac{1}{y} = \frac{3}{\sqrt{2}}$$ $$-(3)$$

From $$(1)$$: $$\frac{1}{x} + \frac{1}{z} = \frac{2}{y}$$. Substituting into $$(3)$$:

$$\frac{2}{y} + \frac{1}{y} = \frac{3}{\sqrt{2}}$$

$$\frac{3}{y} = \frac{3}{\sqrt{2}}$$

$$y = \sqrt{2}$$

From $$(2)$$: $$xz = 2y^2 = 2 \times 2 = 4$$ $$-(4)$$

From $$(1)$$: $$\frac{1}{x} + \frac{1}{z} = \frac{2}{\sqrt{2}} = \sqrt{2}$$, so $$\frac{x + z}{xz} = \sqrt{2}$$

Using $$(4)$$: $$x + z = \sqrt{2} \times 4 = 4\sqrt{2}$$

Therefore: $$x + y + z = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2}$$

$$3(x + y + z)^2 = 3 \times (5\sqrt{2})^2 = 3 \times 50 = 150$$