We have,

Let $AD$ and $BE$ intersect at $P$.
Let $A,B,C,D,E,P$ have position vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d},\overrightarrow{e},\overrightarrow{p}$ respectively.

$D$ and $E$ divide segments $BC$ and $CA$ internally in the ratio $2:1$.

By the section formula for internal division,
$\overrightarrow{d}=\frac{2\times \overrightarrow{c}+1\times \overrightarrow{b}}{2+1}$

$\Rightarrow 3\overrightarrow{d}=2\overrightarrow{c}+\overrightarrow{b} ...\left(1\right)$

And,

$\overrightarrow{e}=\frac{2\overrightarrow{a}+\overrightarrow{c}}{2+1}$

$\Rightarrow 3\overrightarrow{e}=2\overrightarrow{a}+\overrightarrow{c} ....\left(2\right)$

From $\left(1\right)$,

$3\overrightarrow{d}-\overrightarrow{b}=2\overrightarrow{c} .....\left(3\right)$

From $\left(2\right)$,

$3\overrightarrow{e}-2\overrightarrow{a}=\overrightarrow{c}$

$\Rightarrow 2\overrightarrow{c}=6\overrightarrow{e}-4\overrightarrow{a} ....\left(4\right)$

Equating both values of $2\overrightarrow{c}$, we get

$6\overrightarrow{e}-4\overrightarrow{a}=3\overrightarrow{d}-\overrightarrow{b}$

$\Rightarrow 6\overrightarrow{e}+\overrightarrow{b}=4\overrightarrow{a}+3\overrightarrow{d}$

$\Rightarrow \frac{4\overrightarrow{a}+3\overrightarrow{d}}{4+3}=\frac{6\overrightarrow{e}+\overrightarrow{b}}{6+1}$

LHS is the position vector of the point which divides segment $AD$ internally in the ratio $3:4$

RHS is the position vector of the point which divides segment $BE$ internally in the ratio $6:1$.

But $P$ is the point of intersection of $AD$ and $BE$.

$\therefore P$ divides $AD$ internally in the ratio $3:4$.