Triangle $ABC$ has $AB=27$, $BC=25$, and $CA=26$. Let $I$ denote the intersection of the internal angle bisectors of $\triangle ABC$. What is $BI$? $\textbf{(A)}\ 15\qquad$ $\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad$ $\textbf{(C)}\ 3\sqrt{26}\qquad$ $\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad$ $\textbf{(E)}\ 9\sqrt{3}$