442*. Solve the system of equations:

$$
\left\{\begin{array}{l}
x+y^{2}+z^{3}=\frac{13}{6} \\
9 x^{2}+16 y^{4}+144 z^{6}=26
\end{array}\right.
$$
$\triangle$ Vectors $\bar{u}$ and $\bar{v}$ are chosen differently than before. Using the left part of the second equation, we introduce the vector $\bar{u}\left(3 x ; 4 y^{2} ; 12 z^{3}\right)$. Now, we introduce the vector $\bar{v}$ such that $\bar{u} \cdot \bar{v}=x+y^{2}+z^{3}$. The vector $\bar{v}\left(\frac{1}{3} ; \frac{1}{4} ; \frac{1}{12}\right)$ will work. We will have:

$$
\bar{u} \cdot \bar{v}=x+y^{2}+z^{3}=\frac{13}{6} \leq|\bar{u}| \cdot|\bar{v}|=\sqrt{\left(9 x^{2}+16 y^{4}+144 z^{6}\right)\left(\frac{1}{9}+\frac{1}{16}+\frac{1}{144}\right)}=\sqrt{26 \cdot \frac{26}{144}}=\frac{26}{12}=\frac{13}{6}
$$

We obtained that $\bar{u} \cdot \bar{v}=|\bar{u}| \cdot|\bar{v}|$. Therefore,

$$
9 x=16 y^{2}=144 z^{3}, \quad x+y^{2}+z^{3}=\frac{13}{6}
$$

It remains to solve this system of three equations with three unknowns.

Answer: $(4 / 3 ; \pm \sqrt{3} / 2 ; \sqrt[3]{18} / 12)$.