Let $$P$$ be a point in the interior of a triangle $$ABC$$ and let $$AP, BP, CP$$ meet the sides $$BC, CA, AB$$ in $$D, E, F$$ respectively. If $$ \frac{BP}{PE}=\frac{5}{2}, \frac{CP}{PF}=\frac{7}{3},$$ and $$ \frac{AP}{PD}=\frac{p}{q}, $$ where $$p$$ and $$q$$ are natural numbers and $$gcd(𝑝, 𝑞) = 1$$, find $$𝑝 + 𝑞$$.

Assign masses $$m_A , m_B , m_C$$ to the vertices $$A , B , C$$ respectively and use the Mass-Points rule:

If a point $$P$$ lies on the segment joining two points with masses $$m_1 , m_2$$, then
$$\dfrac{\text{distance from }P\text{ to point with mass }m_1}{\text{distance from }P\text{ to point with mass }m_2}= \dfrac{m_2}{m_1}$$.

1. Along $$B \!-\! P \!-\! E$$ we have $$\dfrac{BP}{PE}= \dfrac{5}{2}$$.
 Mass at $$B$$ is $$m_B$$, mass at $$E$$ is the sum of the endpoint masses on $$CA$$, i.e. $$m_E = m_A + m_C$$.
 Therefore $$\dfrac{m_E}{m_B}= \dfrac{5}{2}\;\; \Rightarrow\;\; m_A + m_C = \tfrac{5}{2}\,m_B$$.  $$(1)$$

2. Along $$C \!-\! P \!-\! F$$ we have $$\dfrac{CP}{PF}= \dfrac{7}{3}$$.
 Mass at $$C$$ is $$m_C$$, mass at $$F$$ is the sum on $$AB$$, i.e. $$m_F = m_A + m_B$$.
 Therefore $$\dfrac{m_F}{m_C}= \dfrac{7}{3}\;\; \Rightarrow\;\; m_A + m_B = \tfrac{7}{3}\,m_C$$.  $$(2)$$

Choose $$m_B = 2t$$ so that equation $$(1)$$ becomes $$m_A + m_C = 5t$$.

Put $$m_C = c$$. Then $$m_A = 5t - c$$ and substitute these in $$(2)$$:
$$(5t - c) + 2t = \tfrac{7}{3}\,c \;\; \Rightarrow\;\; 7t - c = \tfrac{7}{3}\,c$$
$$\Rightarrow\; 7t = \tfrac{10}{3}\,c \;\; \Rightarrow\;\; c = \tfrac{21}{10}\,t.$$

Hence $$m_A = 5t - \tfrac{21}{10}t = \tfrac{29}{10}t,\qquad m_B = 2t = \tfrac{20}{10}t,\qquad m_C = \tfrac{21}{10}t.$$

Multiply by $$10$$ to clear the denominator and obtain integral masses:
$$m_A : m_B : m_C = 29t : 20t : 21t.$$ The common factor $$t$$ is irrelevant, so take
$$m_A = 29,\; m_B = 20,\; m_C = 21.$$

3. Point $$D$$ lies on $$BC$$, hence its mass is the sum of the masses at $$B$$ and $$C$$:
$$m_D = m_B + m_C = 20 + 21 = 41.$$

Along $$A \!-\! P \!-\! D$$ we therefore have
$$\dfrac{AP}{PD}= \dfrac{m_D}{m_A}= \dfrac{41}{29}.$$

Thus $$\dfrac{AP}{PD}= \dfrac{p}{q}= \dfrac{41}{29} \; \Rightarrow\; (p,q)=(41,29).$$

Since $$\gcd(41,29)=1$$, we get $$p+q = 41 + 29 = 70.$$

Answer: 70