1. Given non-negative real numbers $x, y, z$ satisfy
$$
x+2 y+3 z=3, \quad 3 x+2 y+z=4 \text {. }
$$

Then the maximum value of $M=3 x^{2}-4 y^{2}+5 z^{2}$ is $\qquad$, and the minimum value is $\qquad$.
2. 1. $\frac{23}{4},-\frac{11}{2}$.

Then $M=3 x^{2}-4 y^{2}+5 z^{2}$
$=3\left(\frac{1}{2}+z\right)^{2}-4\left(\frac{5}{4}-2 z\right)^{2}+5 z^{2}$
$=-8 z^{2}+23 z-\frac{11}{2}$
$=-8\left(z-\frac{23}{16}\right)^{2}+\frac{353}{32}$.
Since $x$, $y$, and $z$ are all non-negative real numbers, we have
$$
\left\{\begin{array}{l}
\frac{1}{2}+z \geqslant 0, \\
\frac{5}{4}-2 z \geqslant 0, \\
z \geqslant 0
\end{array} \Rightarrow 0 \leqslant z \leqslant \frac{5}{8}\right. \text {. }
$$

Therefore, when $z=0$, $M_{\text {min }}=-\frac{11}{2}$;
when $z=\frac{5}{8}$, $M_{\max }=\frac{23}{4}$.