Five. (20 points) Let the binary function be
$$
z=f(x, y)=2 x^{2}+3 y^{2}-6 y
$$

with the domain
$$
D=\left\{(x, y) \mid 3 x^{2}+2 y^{2} \leqslant 7 x y, x, y \in \mathbf{R}\right\} \text {. }
$$
(1) Find the range of values of $z=f(x, y)$ (for $(x, y) \in D$);
(2) Find all real numbers $a$ such that in the Cartesian coordinate system $O-x y z$, the surface $z=f(x, y)$ (for $(x, y) \in D$) intersects with the surface $z=x y+a$ (for $x, y \in \mathbf{R}$).
(1) When $x=0$, $2 y_{-}^{2} \leqslant 0, y=0$,
$$
f(x, y)=f(0,0)=0 \text {; }
$$

When $x \neq 0$, $2\left(\frac{y}{x}\right)^{2}-7 \times \frac{y}{x}+3 \leqslant 0$, that is
$$
\left(\frac{y}{x}-\frac{1}{2}\right)\left(\frac{y}{x}-3\right) \leqslant 0 \text {. }
$$

Solving, we get $\frac{1}{2} \leqslant \frac{y}{x} \leqslant 3$.
$$
\begin{array}{l}
\text { Let } \frac{y}{x}=t \text {, then } \frac{1}{2} \leqslant t \leqslant 3, y=t x, \\
f(x, y)=3 t^{2} x^{2}+2 x^{2}-6 t x \\
=\left(3 t^{2}+2\right) x^{2}-6 t x \\
=x\left[\left(3 t^{2}+2\right) x-6 t\right] .
\end{array}
$$

First, fix $t$ and let $x$ vary.
Clearly, when $x \rightarrow-\infty$ or $+\infty$, $f(x, y) \rightarrow+\infty$.
When $\bar{x}=\frac{3 t}{3 t^{2}+2}$, $f(x, y)$ reaches its minimum value.
$$
\begin{array}{c}
f(x, y)=\frac{-9 t^{2}}{3 t^{2}+2}=-3+\frac{6}{3 t^{2}+2} \\
\geqslant-3+\frac{6}{3^{3}+2}=-\frac{81}{29} .
\end{array}
$$

Equality holds if and only if $t=3, x=\frac{3 t}{3 t^{2}+2}=\frac{9}{29}, y=\angle x=\frac{27}{29}$.

From the above discussion, the range of $f(x, y)$ is $\left[-\frac{81}{29},+\infty\right)$.
(2) The surface $z=f(x, y)((x, y) \in D)$ intersects with $z=x y+a$ $(x, y \in \mathbf{R})$
$\Leftrightarrow$ the equation $f(x, y)=x y+a((x, y) \in D)$ has real solutions
$\Leftrightarrow 2 x^{2}+3 y^{2}-6 y=x y+a((x, y) \in D)$ has real solutions $(x, y)$
$\Leftrightarrow\left\{\begin{array}{l}2 x^{2}+3 t^{2} x^{2}-6 t x=t x^{2}+a, \\ \frac{1}{2} \leqslant t \leqslant 3\end{array}\right.$ has real solutions $(x, t)$
$\Leftrightarrow\left\{\begin{array}{l}\left(3 t^{2}-t+2\right) x^{2}-6 t x-a=0, \\ \frac{1}{2} \leqslant t \leqslant 3\end{array}\right.$ has real solutions $(x, t)$
$\Leftrightarrow\left\{\begin{array}{l}\Delta=36 t^{2}+4\left(3 t^{2}-t+2\right) a \geqslant 0, \\ \frac{1}{2} \leqslant t \leqslant 3\end{array}\right.$ has real solutions $t$
$\Leftrightarrow\left\{\begin{array}{l}a \geqslant \frac{-9 t^{2}}{3 t^{2}-t+2}\left(\text { clearly } 3 t^{2}-t+2>0\right) \\ \frac{1}{2} \leqslant t \leqslant 3\end{array}\right.$,
$\Leftrightarrow a \geqslant-3-3 \cdot \frac{t-2}{3 t^{2}-t+2}\left(\frac{1}{2} \leqslant t \leqslant 3\right)$.
Let $g(t)=\frac{t-2}{3 t^{2}-t+2}\left(\frac{1}{2} \leqslant t \leqslant 3\right)$.
To find the maximum value of $g(t)$, we only need to consider $20$, and it is strictly decreasing with respect to $k$).
Equality holds if and only if $k=1$.
Thus, the maximum value of $g(t)$ is $\frac{1}{26}$.
Therefore, $a \geqslant-3-3 g(t) \geqslant-\frac{81}{26}$.
Hence, the range of $a$ is $a \geqslant-\frac{81}{26}$.