1. Let $x, y, z \in R$ and $x+2 y+3 z=1, y z+z x+x y=-1$, then the range of $x+y+z$ is $\qquad$
1. $\left[\frac{3-3 \sqrt{3}}{4}, \frac{3+3 \sqrt{3}}{4}\right]$

Analysis: Let $x+y+z=t$, since $x+2 y+3 z=1$, we have $\left\{\begin{array}{c}x+y=t-z \\ x+2 y=1-3 z\end{array} \Rightarrow\left\{\begin{array}{l}x=2 t+z-1 \\ y=1-t-2 z\end{array}\right.\right.$;
Thus, $z(t-z)+(2 t+z-1)(1-t-2 z)=-1$,
Simplifying, we get: $3 z^{2}-(3-4 t) z+2 t^{2}-3 t=0$, since the discriminant is not less than 0, hence
$$
\begin{array}{l}
(3-4 t)^{2}-12\left(2 t^{2}-3 t\right) \geq 0 \\
\Leftrightarrow-8 t^{2}+12 t+9 \geq 0 \Leftrightarrow \frac{3-3 \sqrt{3}}{4} \leq t \leq \frac{3+3 \sqrt{3}}{4}
\end{array}
$$