jmc

By Cauchy-Schwarz applied to $\left(1,\frac{1}{3},\frac{1}{2}\right)$ and $(\sqrt{x+27},\sqrt{13-x},\sqrt{x})$, $$\left(1+\frac{1}{3}+\frac{1}{2}\right)((x+27)+3(13-x)+2x)\ge (\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}{)}^2.$$Hence, $$(\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}{)}^2\le 121,$$so $\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}\le 11.$

Equality occurs when $x=9,$ so the maximum value is $\boxed{11}.$