$D,E,F$ are respectively the points on the sides $BC,CA$ and $AB$ of a $∆ABC$ dividing them in the ratio $2:3,1:2,3:1$ internally. The lines $BE$ and $CF$ intersect on the line $AD$ at $P$. If $\overrightarrow{AP}=x_1\overrightarrow{AB}+y_1·\overrightarrow{AC}$, then $x_1+y_1=$

By using section formula we can obtain required result
If vector $\left(d,e,f\right)$ are position vector of points $D, E$ & $F$ respectively then, by section formula

$\overrightarrow{d}=\frac{2\overrightarrow{c}+3\overrightarrow{b}}{5} ...\left(\mathrm{i}\right)$

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

$\overrightarrow{f}=\frac{3\overrightarrow{b}+\overrightarrow{a}}{4} ...\left(\mathrm{iii}\right)$

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Equation of line $AD$ is vector

$\overrightarrow{r}=\overrightarrow{a}+t\left(\overrightarrow{d}-\overrightarrow{a}\right)$

Equation of line $BE$ is

$\overrightarrow{r}=\overrightarrow{b}+m\left(\overrightarrow{e}-\overrightarrow{b}\right)$

For intersection of vector $AD$ and vector $BE$ we need that

"$\overrightarrow{a}+t\left(\overrightarrow{d}-\overrightarrow{a}\right)=\overrightarrow{b}+m\left(\overrightarrow{e}-\overrightarrow{b}\right)$"

true for some $0<t, m<1$

$\overrightarrow{a}+t\frac{\left(2\overrightarrow{c}+3\overrightarrow{b}-5\overrightarrow{a}\right)}{5}=\overrightarrow{b}+m\frac{\left(2\overrightarrow{c}+\overrightarrow{a}-3\overrightarrow{b}\right)}{3}$

$\Rightarrow \left(1-t\right)\overrightarrow{a}+\left(\frac{3t}{5}\right)\overrightarrow{b}+\left(\frac{2t}{5}\right)\overrightarrow{c}=\left(\frac{m}{3}\right)\overrightarrow{a}+\left(1-m\right)\overrightarrow{b}+\left(\frac{2m}{3}\right)\overrightarrow{c}$

Therefore,

$1-t=\frac{m}{3}$

$\frac{3t}{5}=1-m$

$\frac{2t}{5}=\frac{2m}{3}\Rightarrow 3t=5m$

On solving, we get

$t=\frac{{\displaystyle 5}}{6}, m=\frac{{\displaystyle 1}}{2}$

The existence of $t$ and $m$ assures the intersection of vector $AD$ and vector $BE$. The point of intersection is

$\overrightarrow{r}=\overrightarrow{a}+\frac{5}{6}\left(\overrightarrow{d}-\overrightarrow{a}\right)$

$\Rightarrow \overrightarrow{r}=\frac{\left(\overrightarrow{a}+3\overrightarrow{b}+2\overrightarrow{c}\right)}{6}$

Now,

$\overrightarrow{AP}=\overrightarrow{r}-\overrightarrow{a}=\frac{\left(\overrightarrow{a}+3\overrightarrow{b}+2\overrightarrow{c}\right)}{6}-\overrightarrow{a}$

$\Rightarrow \overrightarrow{AP}=\frac{\left(-5\overrightarrow{a}+3\overrightarrow{b}+2\overrightarrow{c}\right)}{6}$

$\Rightarrow \overrightarrow{AP}=\frac{\left(-3\overrightarrow{a}+3\overrightarrow{b}+2\overrightarrow{c}-2\overrightarrow{a}\right)}{6}$

$\Rightarrow \overrightarrow{AP}=\frac{3\left(\overrightarrow{b}-\overrightarrow{a}\right)+2\left(\overrightarrow{c}-\overrightarrow{a}\right)}{6}$

$\Rightarrow \overrightarrow{AP}=\frac{1}{2}\left(\overrightarrow{b}-\overrightarrow{a}\right)+\frac{1}{3}\left(\overrightarrow{c}-\overrightarrow{a}\right)$

$\Rightarrow \overrightarrow{AP}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}$

So, $x_1=\frac{1}{2}, y_1=\frac{1}{3}$

Hence, $x_1+y_1=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$