`@Mon`
`B=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{(\frac{5}{12})-\frac{1}{\sqrt{6})}`
`=\frac{\sqrt{3}}{\sqrt{3}\sqrt{3}}+\frac{2}{3\sqrt{2}\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{3}\sqrt{3}}\sqrt{\frac{5}{12}-\frac{\sqrt{6}}{\sqrt{6}\sqrt{6}}}`
`=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}\sqrt{\frac{5}{12}-\frac{\sqrt{6}}{\sqrt{2}}}`
`=\frac{\sqrt{3}}{3}(1+\sqrt{\frac{5}{12}-\frac{2\sqrt{6}}{12}})+\frac{\sqrt{2}}{6}`
`=\frac{\sqrt{3}}{3}(1+\sqrt{\frac{5-2\sqrt{6}}{12}})+\frac{\sqrt{2}}{6}`
`=\frac{\sqrt{3}}{3}(1+\sqrt{\frac{\sqrt{3^2}-2\sqrt{3}.\sqrt{2}+\sqrt{2^2}}{\sqrt{12}}})+\frac{\sqrt{2}}{6}`
`=\frac{\sqrt{3}}{3}(1+\frac{(\sqrt{3}-\sqrt{2})^2}{\sqrt{12}})+\frac{\sqrt{2}}{6}`
`=\frac{\sqrt{3}}{3}(\frac{\sqrt{12}+\sqrt{3}-\sqrt{2}}{\sqrt{12}})+\frac{\sqrt{2}}{6}`
`=\frac{1}{6}(3\sqrt{3}-\sqrt{2})+\frac{\sqrt{2}}{6}`
`=\frac{3\sqrt{3}-\sqrt{2}+\sqrt{2}}{6}`
`=\frac{3\sqrt{3}}{6}`
`=\frac{\sqrt{3}}{2}`
