For the function $y=f(x)(x \in D)$, if there exists a real number $c(c \in D)$ such that $f(c)=0$ holds, then $x=c$ is called a zero of the function $y=f(x)$. In fact, the zero of the function is the x-coordinate of the intersection point between the function graph and the $x$-axis.
(2007 - Guangdong) It is known that $a$ is a real number, and the function $f(x)=2 a x^{2}+2 x-3-a$. If the function $y=$ $f(x)$ has a zero in the interval $[-1,1]$, find the range of values for $a$.
If $a=0$, then $f(x)=2x-3$, the zero point is $x=\frac{3}{2}$, which does not satisfy the problem's conditions. Therefore, the function $f(x)$ must be a quadratic function, with the discriminant $\Delta=4-8a(-3-a)$, and the axis of symmetry is $x=-\frac{1}{2a}$. Thus, we should have
$$
f(1) f(-1) \leqslant 0 \text { or }\left\{\begin{array}{l}
-1 \leqslant-\frac{1}{2a} \leqslant 1, \\
\Delta \geqslant 0, \\
f(1) f(-1) \geqslant 0,
\end{array}\right.
$$

which means
$$
(2a+2-3-a)(2a-2-3-a) \leqslant 0 \text { or }\left\{\begin{array}{l}
a \leqslant-\frac{1}{2} \text { or } a \geqslant \frac{1}{2}, \\
2a^{2}+6a+1 \geqslant 0, \\
(2a+2-a-3)(2a-2-3-a) \geqslant 0,
\end{array}\right.
$$

Solving this, we get $1 \leqslant a \leqslant 5$ or $a \leqslant \frac{-3-\sqrt{7}}{2}$ or $a \geqslant 5$. Therefore, the range of $a$ is $a \leqslant \frac{-3-\sqrt{7}}{2}$ or $a \geqslant 1$.

For the function $y=f(x)(x \in D)$, if for any $x_{1}, x_{2} \in D$ there is always $\frac{1}{2}\left[f\left(x_{1}\right)+f\left(x_{2}\right)\right] \leqslant f\left(\frac{x_{1}+x_{2}}{2}\right)$, then this function is called a convex function; if $\frac{1}{2}\left[f\left(x_{1}\right)+f\left(x_{2}\right)\right] \geqslant f\left(\frac{x_{1}+x_{2}}{2}\right)$, then this function is called a concave function.
The "convexity" of a function is a description of the curvature of its graph.
It is known that a concave function has the following property: for any $x_{i} \in(a, b)(i=1,2, \cdots, n)$, there must be
$$
f\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) \leqslant \frac{f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{n}\right)}{n}
$$

where the equality holds if and only if $x_{1}=x_{2}=\cdots=x_{n}$.
(1) Prove that $y=x^{2}$ is a concave function on $(-\infty,+\infty)$;
(2) If $x, y, z \in \mathbf{R}$, and $x+2y+3z=12$, try to find the minimum value of $x^{2}+2y^{2}+3z^{2}$ and the values of $x, y, z$ when the minimum value is achieved.
Proof (1) $\frac{1}{2}\left(x_{1}^{2}+x_{2}^{2}\right)-\left(\frac{x_{1}+x_{2}}{2}\right)^{2}=\frac{1}{4}\left(x_{1}-x_{2}\right)^{2} \geqslant 0$,
thus
$$
\frac{1}{2}\left(x_{1}^{2}+x_{2}^{2}\right) \geqslant\left(\frac{x_{1}+x_{2}}{2}\right)^{2},
$$

Therefore, the function $y=x^{2}$ is a concave function.
(2) By the property of concave functions, we have
$$
\frac{1}{6}\left(x^{2}+y^{2}+y^{2}+z^{2}+z^{2}+z^{2}\right) \geqslant\left(\frac{x+y+y+z+z+z}{6}\right)^{2}=4,
$$

which means $x^{2}+2y^{2}+3z^{2} \geqslant 24$. Therefore, when $x=y=z=2$, $x^{2}+2y^{2}+3z^{2}$ achieves its minimum value of 24.