Three. (Full marks 20 points) Let real numbers $x^{2}+y^{2} \leqslant 5$. Find the maximum and minimum values of $f(x, y)=3|x+y|+|4 y+9|+|7 y-3 x-18|$.
Three, let $x=r \cos \theta, y=r \sin \theta(r>0,0 \leqslant \theta < 2\pi)$.
$$
\begin{array}{l}
7 y-3 x-18=7 r \sin \theta-3 r \cos \theta-18 \\
=r \sqrt{7^{2}+3^{2}} \sin (\theta-\alpha)-18 \leqslant \sqrt{5} \times \sqrt{58}-18 \\
=\sqrt{290}-18<0 . \\
\begin{aligned}
\therefore f(x, y)= & 3|x+y|+4 y+9-7 y+3 x+18 \\
= & 3|x+y|+3(x-y)+27 \\
= & 3|r \cos \theta+r \sin \theta|+3(r \cos \theta \\
& -r \sin \theta)+27 \\
= & 3 \sqrt{2} r\left|\sin \left(\theta+\frac{\pi}{4}\right)\right| \\
& +3 \sqrt{2} r \sin \left(\frac{\pi}{4}-\theta\right)+27 .
\end{aligned}
\end{array}
$$
(1) When $\sin \left(\theta+\frac{\pi}{4}\right) \geqslant 0$ (i.e., $0 \leqslant \theta \leqslant \frac{3}{4} \pi$ or $\frac{7 \pi}{4} \leqslant \theta < 2 \pi$),
$$
\begin{aligned}
f(x, y)= & 3 \sqrt{2} r\left[\sin (\theta+\pi) \cdots \sin \left(\theta-\frac{\pi}{4}\right)\right] \\
& +27 \\
= & 3 \sqrt{2} r \cdot 2 \cos \theta \cdot \sin \frac{\pi}{4}+27 \\
= & 6 r \cos \theta+27 .
\end{aligned}
$$

At this time, the maximum value of $\cos \theta$ is $1$ (when $\theta=0$), and the minimum value is $-\frac{\sqrt{2}}{2}$ (when $\theta=\frac{3 \pi}{4}, \frac{7 \pi}{4}$), hence
$$
\begin{array}{l}
f(x, y)_{\max }=6 \times \sqrt{5} \times 1+27=27+6 \sqrt{5}, \\
f(x, y)_{\min }=6 \times \sqrt{5}\left(-\frac{\sqrt{2}}{2}\right)+27=27-3 \sqrt{10} .
\end{array}
$$
(2) When $\sin \left(\theta+\frac{\pi}{4}\right) \leqslant 0$ (i.e., $\frac{3 \pi}{4} \leqslant \theta \leqslant \frac{7 \pi}{4}$),
$$
\begin{aligned}
f(x, y) & =3 \sqrt{2} r\left[\sin \left(\frac{\pi}{4}-\theta\right)-\sin \left(\frac{\pi}{4}+\theta\right)\right]+27 \\
& =3 \sqrt{2} r \cdot 2 \cos \frac{\pi}{4} \cdot \sin (-\theta)+27 \\
& =-6 r \sin \theta+27 .
\end{aligned}
$$

At this time, the maximum value of $\sin \theta$ is $\frac{\sqrt{2}}{2}$ (when $\theta=\frac{3 \pi}{4}$), and the minimum value is $-1$ (when $\theta=\frac{3 \pi}{2}$). Hence
$$
\begin{array}{l}
f(x, y)_{\max }=-6 \times \sqrt{5}(-1)+27=27+6 \sqrt{5}, \\
f(x, y)_{\min }=-6 \times \sqrt{5} \times \frac{\sqrt{2}}{2}+27=27-3 \sqrt{10} .
\end{array}
$$

In summary, $f(x, y)_{\text {max }}=27+6 \sqrt{5}$,
$$
f(x, y)_{\text {min }}=27-3 \sqrt{10} \text {. }
$$