1.102 If $4(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2})=x+y+z+9$, then the value of $x y z$ is
(A) 0.
(B) 2.
(C) 15.
(D) 120.
(Nanchang City, Jiangxi Province, China Junior High School Mathematics Competition, 1991)
［Solution］From
$$
\begin{aligned}
& x+y+z+9 \\
= & (\sqrt{x})^{2}+(\sqrt{y-1})^{2}+(\sqrt{z-2})^{2}+12
\end{aligned}
$$

the given equation can be transformed into
$$
\begin{array}{l}
(\sqrt{x}-2)^{2}+(\sqrt{y-1}-2)^{2}+(\sqrt{z-2}-2)^{2}=0 \text {. } \\
\therefore \quad \sqrt{x}=2, \quad \sqrt{y-1}=2, \quad \sqrt{z-2}=2 \text {, } \\
x=4, \quad y=5, \quad z=6 . \\
x y z=120 \text {. } \\
\end{array}
$$

Therefore, the answer is $(D)$.