7. The real solution $(x, y, z)=$ $\qquad$ of the equation $2 \sqrt{x-4}+3 \sqrt{y-9}+4 \sqrt{z-16}=\frac{1}{2}(x+y+z)$
Solution: The original equation can be transformed into
$$
(x-4+4-4 \sqrt{x-4})+(y-9+9-6 \sqrt{y-9})+(z-16+16-8 \sqrt{z-16})=0 .
$$

which is $(\sqrt{x-4}-2)^{2}+(\sqrt{y-9}-3)^{2}+(\sqrt{z-16}-4)^{2}=0$.
Therefore, $(x, y, z)=(8,18,32)$.
Note: The original problem is:
$\sqrt{x-4}+3 \sqrt{y-9}+4 \sqrt{z-16}=\frac{1}{2}(x+y+z)$, which leads to
$(\sqrt{x-4}-1)^{2}+(\sqrt{y-9}-3)^{2}+(\sqrt{z-16}-4)^{2}+3=0$, with no real solutions.