Subtract the sum of $\frac{2}{3}$ and $\frac{1}{2}$ from the sum of $\frac{5}{6}$, $\frac{7}{12}$ and $\frac{4}{15}$.

Step 1:

Sum of $\frac{2}{3}$ and $\frac{1}{2}$

$\frac{2}{3}+\frac{1}{2}$

$=(\frac{2}{3}\times \frac{2}{2})+(\frac{1}{2}\times \frac{3}{3})$

$=\frac{4}{6}+\frac{3}{6}$

$=\frac{4+3}{6}$

$=\frac{7}{6}$

$\Rightarrow \frac{2}{3}+\frac{1}{2}=\frac{7}{6}$ ---(1)

Step 2:

Sum of $\frac{5}{6},\frac{7}{12}$ and $\frac{4}{15}$

$\frac{5}{6}+\frac{7}{12}+\frac{4}{15}$

LCM of $6,12$ and $15$ is $60$

$=(\frac{5}{6}\times \frac{10}{10})+(\frac{7}{12}\times \frac{5}{5})+(\frac{4}{15}\times \frac{4}{4})$

$=\frac{50}{60}+\frac{35}{60}+\frac{16}{60}$

$=\frac{50+35+16}{60}$

$=\frac{101}{60}$

$\Rightarrow \frac{5}{6}+\frac{7}{12}+\frac{4}{15}=\frac{101}{60}$ ----(2)

Now, lets do (2)-(1)

$\frac{101}{60}-\frac{7}{6}$

$=\frac{101}{60}-(\frac{7}{6}\times \frac{10}{10})$

$=\frac{101}{60}-\frac{70}{60}$

$=\frac{101-70}{60}$

$=\frac{31}{60}$

Hence, the difference of the sum of $\frac{2}{3}$ and $\frac{1}{2}$from the sum of $\frac{5}{6}$, $\frac{7}{12}$ and $\frac{4}{15}$ is $\frac{31}{60}$.