In $\triangle ABC$ lines $CE$ and $AD$ are drawn so that 
$\dfrac{CD}{DB}=\dfrac{3}{1}$ and $\dfrac{AE}{EB}=\dfrac{3}{2}$. Let $r=\dfrac{CP}{PE}$ 
where $P$ is the intersection point of $CE$ and $AD$. Then $r$ equals:

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \dfrac{3}{2}\qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \dfrac{5}{2}$
Draw line $PB$, and let $[PEB] = 2b$, $[PDB] = a$, and $[CAP] = c$, so $[CPD] = 3a$ and $[APE] = 3b$.  Because $\triangle CAE$ and $\triangle CEB$ share an altitude,
\[c + 3b = \tfrac{3}{2} (3a+a+2b)\]
\[c + 3b = 6a + 3b\]
\[c  = 6a\]
Because $\triangle ACD$ and $\triangle ABD$ share an altitude,
\[6a+3a = 3(a+2b+3b)\]
\[9a = 3a+15b\]
\[6a = 15b\]
\[a = \tfrac{5}{2}b\]
Thus, $[CAP] = 15b$, and since $[APE] = 3b$, $r = \tfrac{CP}{PE} = 5$, which is answer choice $\boxed{\textbf{(D)}}$.