1. Let $\sqrt{3+\sqrt{2}+\sqrt{3}+\sqrt{6}}=\sqrt{x}+\sqrt{y}+\sqrt{z}$, and $x, y, z$ are rational numbers. Then $x y z=(\quad)$.
(A) $\frac{3}{4}$
(B) $\frac{5}{6}$
(C) $\frac{7}{12}$
(D) $\frac{13}{18}$
1. A.

Squaring both sides, we get
$$
3+\sqrt{2}+\sqrt{3}+\sqrt{6}=x+y+z+2 \sqrt{x y}+2 \sqrt{y z}+2 \sqrt{x z} \text{. }
$$

Based on the symmetry of the rational numbers $x, y, z$, we can consider the system of equations
$$
\left\{\begin{array}{l}
x+y+z=3, \\
2 \sqrt{x y}=\sqrt{2}, \\
2 \sqrt{y z}=\sqrt{3}, \\
2 \sqrt{x z}=\sqrt{6} .
\end{array}\right.
$$

Solving, we get $x=1, y=\frac{1}{2}, z=\frac{3}{2}$. At this point, $x y z=\frac{3}{4}$.