Solution. Let $a=3 x, b=2 y, z=c$ We then obtain
$$a+b+c=3 x+2 y+z, a^{2}+3 b^{2}+15 c^{2}=a b c .$$

According to the weighted AM-GM inequality, we have that
$$\begin{array}{c} a+b+c \geq(2 a)^{1 / 2}(3 b)^{1 / 3}(6 c)^{1 / 6} \\ a^{2}+3 b^{2}+15 c^{2} \geq\left(4 a^{2}\right)^{1 / 4}\left(9 b^{2}\right)^{3 / 9}\left(36 c^{2}\right)^{15 / 36}=\left(4 a^{2}\right)^{1 / 4}\left(9 b^{2}\right)^{1 / 3}\left(36 c^{2}\right)^{5 / 12} \end{array}$$

Multiplying the results above, we obtain
$$(a+b+c)\left(a^{2}+3 b^{2}+15 c^{2}\right) \geq 36 a b c \Rightarrow a+b+c \geq 36$$

So the minimum of $3 x+2 y+z$ is 36, attained for $x=y=z=6$.