1. Ternary function
$$
\begin{array}{l}
f(x, y, z) \\
=x^{2}+2 y^{2}+3 z^{2}+2 x y+4 y z+2 z x- \\
\quad 6 x-10 y-12 z
\end{array}
$$

The minimum value of the function is $\qquad$
- 1. -14 .

Notice,
$$
\begin{array}{l}
f(x, y, z) \\
=(x+y+z-3)^{2}+(y+z-2)^{2}+(z-1)^{2}-14 \\
\geqslant-14 .
\end{array}
$$

When and only when $x=y=z=1$, the equality holds. Therefore, the minimum value of $f(x, y, z)$ is -14.