11. D.

Notice,
$$
\begin{array}{l}
|\overrightarrow{B P}|^{2}=\left(\frac{2 \overrightarrow{B A}+\overrightarrow{B C}}{3}\right)^{2} \\
= \frac{4|\overrightarrow{B A}|^{2}+|\overrightarrow{B C}|^{2}+4 \overrightarrow{B A} \cdot \overrightarrow{B C}}{9} \\
=\frac{4|\overrightarrow{B A}|^{2}+|\overrightarrow{B C}|^{2}+4|\overrightarrow{B A}||\overrightarrow{B C}| \cos \angle A B C}{9} \\
=\frac{1}{9}(4 \times 169+196+ \\
= \frac{6 \times 169+3 \times 196-2 \times 225}{9} \\
= \frac{1152}{9} .
\end{array}
$$

Thus, $B P=\frac{24 \sqrt{2}}{3}$.
$$
\text { By } \triangle A P D \backsim \triangle C P B
$$
$$
\begin{array}{l}
\Rightarrow \frac{P D}{P B}=\frac{P A}{P C}=\frac{1}{2} \\
\Rightarrow P D=4 \sqrt{2} .
\end{array}
$$

Similarly, $P E=16 \sqrt{2}$.
Therefore, $D E=P E-P D=12 \sqrt{2}$.
11. As shown in Figure 5, in $\triangle A B C$, $A B=13$, $B C=14$, $A C=15$. If $P$ is a point on $A C$ such that $P C=10$, and there exist points $D$ and $E$ on line $B P$ such that quadrilaterals $A B C D$ and $A B C E$ are both trapezoids, then the length of $D E$ is ( ).
(A) $\frac{42}{5}$
(B) $6 \sqrt{2}$
(C) $\frac{84}{5}$
(D) $12 \sqrt{2}$
(E) 18