In triangle $ABC,$ $D$ is on $\overline{AB}$ and $E$ is on $\overline{BC}$ such that $AD = 3,$ $BD = 2,$ $BE = 3,$ and $CE = 4.$  Let $F$ be the intersection of $\overline{AE}$ and $\overline{CD}.$  Find $\frac{EF}{FA}.$
Let $\mathbf{a}$ denote $\overrightarrow{A},$ etc.  Since $AD:BD = 3:2,$
\[\mathbf{d} = \frac{2}{5} \mathbf{a} + \frac{3}{5} \mathbf{b}.\]Since $BE:CE = 3:4,$
\[\mathbf{e} = \frac{4}{7} \mathbf{b} + \frac{3}{7} \mathbf{c}.\][asy]
unitsize(0.6 cm);

pair A, B, C, D, E, F;

A = 5*dir(70);
B = (0,0);
C = (7,0);
D = 2*dir(70);
E = (3,0);
F = extension(A,E,C,D);

draw(A--B--C--cycle);
draw(A--E);
draw(C--D);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, W);
label("$E$", E, S);
label("$F$", F, NE);
label("$3$", (A + D)/2, W, red);
label("$2$", (B + D)/2, W, red);
label("$3$", (B + E)/2, S, red);
label("$4$", (C + E)/2, S, red);
[/asy]

Isolating $\mathbf{b}$ in each equation, we obtain
\[\mathbf{b} = \frac{5 \mathbf{d} - 2 \mathbf{a}}{3} = \frac{7 \mathbf{e} - 3 \mathbf{c}}{4}.\]Then $20 \mathbf{d} - 8 \mathbf{a} = 21 \mathbf{e} - 9 \mathbf{c},$ so $8 \mathbf{a} + 21 \mathbf{e} = 9 \mathbf{c} + 20 \mathbf{d},$ or
\[\frac{8}{29} \mathbf{a} + \frac{21}{29} \mathbf{e} = \frac{9}{29} \mathbf{c} + \frac{20}{29} \mathbf{d}.\]Since the coefficients on both sides of the equation add up to 1, the vector on the left side lies on line $AE,$ and the vector on the right side lies on line $CD.$  Therefore, this common vector is $\mathbf{f}.$  Furthermore, $\frac{EF}{FA} = \boxed{\frac{8}{21}}.$