jmc

Since $\frac{1}{7}<\frac{1}{4}$ and $\frac{1}{19}<\frac{1}{4}$, $$2\frac{1}{7}+3\frac{1}{2}+5\frac{1}{19}<2\frac{1}{4}+3\frac{1}{2}+5\frac{1}{4}=11$$ Clearly, $$2\frac{1}{7}+3\frac{1}{2}+5\frac{1}{19}>2+3\frac{1}{2}+5=10\frac{1}{2}$$ Thus, the sum is between $10\frac{1}{2}$ and $11$. $\boxed{10\frac{1}{2}\text{ and }11}$