(14) If real numbers $x, y, z$ satisfy $x+3 y+2 z=6$, solve the inequality
$$
x^{2}+9 y^{2}-2 x-6 y+4 z \leqslant 8 \text {. }
$$
(14) Method 1: From $x+3 y+2 z=6$, we have $x+3 y=2(3-z)$, thus $x, 3-z, 3 y$ form an arithmetic sequence. Let
$$
\left\{\begin{array}{l}
x=3-z+d \\
3 y=3-z-d
\end{array}\right.
$$

Then
$$
\begin{array}{l}
8 \geqslant x^{2}+9 y^{2}-2 x-6 y+4 z \\
=[(3-z)+d]^{2}+[(3-z)-d]^{2}-2(3-z+d)-2(3-z-d)+4 z \\
\quad=2(3-z)^{2}+2 d^{2}-6+2 z-2 d-6+2 z+2 d+4 z \\
\quad=2 z^{2}-12 z+18+2 d^{2}-12+8 z \\
\quad=2 z^{2}-4 z+2 d^{2}+6 \\
\quad=2(z-1)^{2}+2 d^{2}+4
\end{array}
$$

Thus, $2 \geqslant(z-1)^{2}+d^{2}$, hence
$$
\begin{array}{l}
|d| \leqslant \sqrt{2-(z-1)^{2}}, \\
\sqrt{2}-1 \leqslant z \leqslant \sqrt{2}+1 .
\end{array}
$$

Noting that
$$
\left\{\begin{array}{l}
x=3-z+d, \\
y=2-\frac{2}{3} z-\frac{1}{3} x,
\end{array}\right.
$$

we have
$$
\left\{\begin{array}{l}
\sqrt{2}-1 \leqslant z \leqslant \sqrt{2}+1, \\
3-z-\sqrt{2-(z-1)^{2}} \leqslant x \leqslant 3-z+\sqrt{2-(z-1)^{2}}, \\
y=2-\frac{2}{3} z-\frac{1}{3} x,
\end{array}\right.
$$

Thus, the solution set is
$$
\begin{array}{l}
\left\{(x, y, z) \left\lvert\, y=2-\frac{2}{3} z-\frac{1}{3} x\right.\right. \\
3-z-\sqrt{2-(z-1)^{2}} \leqslant x \leqslant 3-z+\sqrt{2-(z-1)^{2}}, \\
\sqrt{2}-1 \leqslant z \leqslant \sqrt{2}+1\} .
\end{array}
$$

Method 2: Substituting $2 z=6-x-3 y$ into the inequality, we get
$$
x^{2}+9 y^{2}-2 x-6 y+2(6-x-3 y) \leqslant 8,
$$

which simplifies to
$$
(x-2)^{2}+(3 y-2)^{2} \leqslant 4 .
$$

Thus, the solution set is
$$
\left\{(x, y, z) \left\lvert\, z=3-\frac{1}{2} x-\frac{3}{2} y\right.,(x-2)^{2}+(3 y-2)^{2} \leqslant 4\right\},
$$

or
$$
\begin{aligned}
\{(x, y, z) \mid z & =3-\frac{1}{2} x-\frac{3}{2} y, \frac{2}{3}-\frac{1}{3} \sqrt{4-(x-2)^{2}} \\
& \left.\leqslant y \leqslant \frac{2}{3}+\frac{1}{3} \sqrt{4-(x-2)^{2}}, 0 \leqslant x \leqslant 4\right\} .
\end{aligned}
$$