[tex]\frac{1}{4}\sqrt{48}+\frac{1}{2}\sqrt{243}-\frac{1}{6}\sqrt{12} \\
\\
\frac{1}{4}.4\sqrt3+\frac{1}{2}.9\sqrt3-\frac{1}{6}.2\sqrt3= \\
\\
\sqrt3+\frac{9}{2} \sqrt3-\frac{1}{3}\sqrt3= \\
\\
(1+\frac{9}{2}-\frac{1}{3})\sqrt3 = \\
\\
\frac{6+27-2}{6}\sqrt3=\boxed{\frac{31}{6}\sqrt3}[/tex]
