1. Find all triples of real numbers $(x, y, z)$ such that
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+25 z^{2}=6 x z+8 y z, \\
3 x^{2}+2 y^{2}+z^{2}=240 .
\end{array}\right.
$$
1. Note that,

Equation (1)
$$
\begin{array}{l}
\Leftrightarrow x^{2}-6 x z+9 z^{2}+y^{2}-8 y z+16 z^{2}=0 \\
\Leftrightarrow(x-3 z)^{2}+(y-4 z)^{2}=0 \\
\quad \Leftrightarrow x-3 z=0, y-4 z=0 \\
\quad \Leftrightarrow x=3 z, y=4 z .
\end{array}
$$

Thus, all triples satisfying Equation (1) are
$$
(x, y, z)=(3 t, 4 t, t)(t \in \mathbf{R}),
$$

Substituting them into Equation (2) yields
$$
\begin{array}{l}
3 \times 9 t^{2}+2 \times 16 t^{2}+t^{2}=240 \\
\Rightarrow 60 t^{2}=240 \Rightarrow t^{2}=4 . \\
\Rightarrow t= \pm 2 .
\end{array}
$$

When $t=2$, $(x, y, z)=(6,8,2)$;
When $t=-2$, $(x, y, z)=(-6,-8,-2)$. Therefore, $(x, y, z)=(-6,-8,-2),(6,8,2)$.