jmc

Let $a=11$, $b=13$, and $c=17$. Using these variables the expression becomes $$\frac{a^2\left(\frac{1}{b}-\frac{1}{c}\right)+b^2\left(\frac{1}{c}-\frac{1}{a}\right)+c^2\left(\frac{1}{a}-\frac{1}{b}\right)}{a\left(\frac{1}{b}-\frac{1}{c}\right)+b\left(\frac{1}{c}-\frac{1}{a}\right)+c\left(\frac{1}{a}-\frac{1}{b}\right)}.$$By grouping all the terms with the same reciprocal together we get $$\frac{\frac{1}{a}(c^2-b^2)+\frac{1}{b}(a^2-c^2)+\frac{1}{c}(b^2-a^2)}{\frac{1}{a}(c-b)+\frac{1}{b}(a-c)+\frac{1}{c}(b-a)}.$$Using the difference of squares, we can rewrite the numerator of the expression as $$\frac{1}{a}(c+b)(c-b)+\frac{1}{b}(a+c)(a-c)+\frac{1}{c}(b+a)(b-a).$$Let $S=a+b+c$. Then the numerator is $$\begin{array}{rl}& \frac{1}{a}(S-a)(c-b)+\frac{1}{b}(S-b)(a-b)+\frac{1}{c}(S-c)(b-a)\\ & =\frac{1}{a}(c-b)S-(c-b)+\frac{1}{b}(a-b)S-(a-c)+\frac{1}{c}(b-a)S-(b-a)\\ & =\left[\frac{1}{a}(c-b)+\frac{1}{b}(a-b)+\frac{1}{c}(b-a)\right]S\end{array}$$But this is just the denominator of our fraction times $S$. So our original expression simplifies to $S$ which is $a+b+c=11+13+17=\boxed{41}$.