Place $A$ at the origin and use vectors. Let $\vec b = \vec{AB}, \vec c = \vec{AC}.$

$D$ on $BC$ with $BD:DC=2:3$: $\vec{AD} = \dfrac{3\vec b + 2\vec c}{5}.$

$E$ on $AD$ with $AE:ED=3:2$: $\vec{AE} = \dfrac{3}{5}\vec{AD} = \dfrac{3(3\vec b + 2\vec c)}{25} = \dfrac{9\vec b + 6\vec c}{25}.$

Parametrise line $BE$: $\vec r = \vec b + t(\vec{AE} - \vec b) = \vec b + t\!\left(\dfrac{9\vec b + 6\vec c}{25} - \vec b\right) = \vec b\!\left(1 - \dfrac{16t}{25}\right) + \vec c\cdot\dfrac{6t}{25}.$

For this point to lie on line $AC$ (i.e., $\vec r = s\vec c$), the coefficient of $\vec b$ must vanish:

$

1 - \dfrac{16t}{25} = 0 ;\Longrightarrow; t = \dfrac{25}{16}.

$

Then $\vec r = \vec c\cdot\dfrac{6}{25}\cdot\dfrac{25}{16} = \dfrac{3}{8}\vec c.$

So $\vec{AF} = \dfrac{3}{8}\vec c$, meaning $AF:AC = 3:8$, hence

$

AF:FC = 3:5.

$

Cross-check by mass points: Place masses: at $C$ mass $2$, at $B$ mass $3$ so $D$ has mass $5$ (since $BD:DC=2:3$). Place mass at $A$ such that $AE:ED=3:2$, i.e., mass at $A$ should be in ratio $2:3$ with mass at $D$, so mass at $A$ is $2$ and mass at $D$ is $3$ (consistent up to scale; rescale $D$'s mass from $5$ to $3$ by dividing by $5/3$; equivalently, multiply $A$'s mass by $5/3$). To keep arithmetic clean: mass at $A=2\cdot 5 = 10$, at $D = 3\cdot 5 = 15$, so at $B$ effective mass $= (3/5)\cdot 15 = 9$ and at $C$ effective mass $= (2/5)\cdot 15 = 6.$ Then on $AC$, masses $10$ at $A$ and $6$ at $C$, so $F$ divides $AC$ in ratio $6:10 = 3:5$ from $A$. \checkmark

Why this is CAT-level: The natural temptation is to use Menelaus on $\triangle ADC$ with transversal $B$--$E$--$F$, but choosing the right triangle and getting the signed ratios correct is error-prone under time pressure. Vector / mass-point methods are faster and trap-resistant. Wrong option (C) $9:10$ corresponds to a common Menelaus sign error.

Answer: (A) $3:5$.