HMMT February

Solution:
Let $k = \log_{2} 3$ for brevity. Taking the base-$2$ log of each equation gives
$$(\log_{2} x)(\log_{2} y + \log_{2} z) = 8 + 4k,$$
$$(\log_{2} y)(\log_{2} z + \log_{2} x) = 9 + 6k,$$
$$(\log_{2} z)(\log_{2} x + \log_{2} y) = 5 + 10k.$$
Adding the first two equations and subtracting the third yields $2 \log_{2} x \log_{2} y = 12$, so $\log_{2} x \log_{2} y = 6$. Similarly, we get
$$\log_{2} x \log_{2} y = 6,$$
$$\log_{2} y \log_{2} z = 3 + 6k,$$
$$\log_{2} z \log_{2} x = 2 + 4k.$$
Multiplying the first two equations and dividing by the third yields $(\log_{2} y)^{2} = 9$, so $\log_{2} y = \pm 3$. Then, the first and last equations tell us $\log_{2} x = \pm 2$ and $\log_{2} z = \pm (1 + 2k)$, with all signs matching. Thus
$$\log_{2} x + \log_{2} y + \log_{2} z = \pm (3 + 2 + (1 + 2k)) = \pm (6 + 2k),$$
so
$$x y z = 2^{\pm (6 + 2k)} = 2^{6} \cdot 3^{2} \quad \text{or} \quad 2^{-6} \cdot 3^{-2}.$$
Clearly, the smallest solution is $2^{-6} \cdot 3^{-2} = \left[\frac{1}{576}\right]$.