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By Cauchy-Schwarz, $$(1+1+1)(3x+1+3y+3+3z+5)\ge (\sqrt{3x+1}+\sqrt{3y+3}+\sqrt{3z+5}{)}^2.$$Then $$(\sqrt{3x+1}+\sqrt{3y+3}+\sqrt{3z+5}{)}^2\le (3)(3+1+3+5)=36,$$so $\sqrt{3x+1}+\sqrt{3y+3}+\sqrt{3z+5}\le 6.$

Equality occurs when $3x+1=3y+3=3z+5.$ Along with the condition $x+y+z=1,$ we can solve to get $x=1,$ $y=\frac{1}{3},$ $z=-\frac{1}{3}.$ Thus, the maximum value is $\boxed{6}.$