By denoting $\frac{1}{x\sqrt{2}} = a$, $\frac{1}{y\sqrt{3}} = b$, $\frac{1}{2z} = c$, the problem becomes:
Find the greatest value of the expression
$$Q(a, b, c) = 2a^2 + 6b^2 + 12c^2$$
where $a$, $b$, $c$ are positive numbers satisfying the conditions:
$$\max\{a, b\} < c \leq \frac{1}{\sqrt{2}} \quad (1)$$
$$c\sqrt{2} + a\sqrt{3} \geq 2\sqrt{6} \text{ ac.} \quad (2)$$
$$c\sqrt{2} + b\sqrt{5} \geq 2\sqrt{10} \text{ bc.} \quad (3)$$
We have from (2):
$$\frac{\sqrt{2}}{a} + \frac{\sqrt{3}}{c} \geq 2\sqrt{6} \Rightarrow \frac{2}{a^2} + \frac{3}{c^2} \geq 12 \Rightarrow \frac{1}{6}a^2\left(\frac{2}{a^2} + \frac{3}{c^2}\right) \geq 2a^2,$$
$$\text{hence } a^2 + c^2 = 2a^2 + c^2 - a^2 \leq \frac{1}{6}a^2\left(\frac{2}{a^2} + \frac{3}{c^2}\right) + c^2\left(1 - \frac{a^2}{c^2}\right) \leq \frac{1}{6}a^2\left(\frac{2}{a^2} + \frac{3}{c^2}\right) + \frac{1}{2}\left(1 - \frac{a^2}{c^2}\right) = \frac{5}{6}.$$
Analogously, from (1) and (3) we have: $b^2 + c^2 \leq \frac{7}{10}$.
Thus,
$$Q(a, b, c) = 2(a^2 + c^2) + 6(b^2 + c^2) + 4c^2 \leq \frac{118}{15}.$$
It is easy to verify that $Q(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{5}}, \frac{1}{\sqrt{2}}) = \frac{118}{15}$ and the values $a = \frac{1}{\sqrt{3}}$, $b = \frac{1}{\sqrt{5}}$, $c = \frac{1}{\sqrt{2}}$ satisfy the conditions (1) - (2) - (3).
$$\text{Conclusion: } \max P(x, y, z) = \max Q(a, b, c) = \frac{118}{15}.$$