Let the numbers $x, y$ satisfy $x^{2}+y^{2} \leqslant 5$. Then the maximum value of the algebraic expression $3|x+y|+|4 y+9!+| 7 y-3 x-18 \mid$ is $\qquad$ .
10. $27+6 \sqrt{5}$.

Since the distances from the point $(0,0)$ to the lines $4 y+9=0$ and $7 y-3 x-18=0$ are $\frac{9}{4}$ and $\frac{18}{\sqrt{58}}$, respectively, both greater than $\sqrt{5}$, we have $|4 y+9|+|7 y-3 x-18|=4 y+9+18+3 x-7 y=27+3(x-y)$. Therefore, $S=3|x+y|$ $\top 27+3(x-y)=27+3(|x+y|+(x-y)) \leqslant 27+6 \sqrt{\frac{(x+y)^{2}+(x-y)^{2}}{2}} \leqslant 27+6 \sqrt{5}$. Equality is achieved when $x=\sqrt{5}, y=0$.