Step 1: Understanding the Concept:
This problem can be elegantly solved using the concept of Mass Point Geometry or by using position vectors. We will use mass point geometry for a simpler calculation.
We assign masses to the vertices of the triangle such that the center of mass lies at the intersection point \( G \).
Step 2: Key Formula or Approach:
If a point \( X \) on segment \( YZ \) divides it in ratio \( m:n \) (i.e., \( YX:XZ = m:n \)), we can assign mass \( n \) at \( Y \) and mass \( m \) at \( Z \). The mass at \( X \) will be \( m+n \).
Step 3: Detailed Explanation:
Given \( P \) divides \( BC \) in ratio \( 3:4 \), so \( BP:PC = 3:4 \).
Assign mass \( m_B \) at \( B \) and \( m_C \) at \( C \) such that \( m_B \cdot BP = m_C \cdot PC \).
\( m_B \cdot 3 = m_C \cdot 4 \implies \frac{m_B}{m_C} = \frac{4}{3} \).
Let's choose \( m_B = 4 \) and \( m_C = 3 \).
The mass at \( P \) will be \( m_P = m_B + m_C = 4 + 3 = 7 \).
Given \( Q \) divides \( CA \) internally in the ratio \( 5:3 \). Standard convention implies \( CQ:QA = 5:3 \).
Assign mass \( m_C \) at \( C \) and \( m_A \) at \( A \) such that \( m_C \cdot CQ = m_A \cdot QA \).
We already have \( m_C = 3 \).
\( 3 \cdot 5 = m_A \cdot 3 \implies 15 = 3m_A \implies m_A = 5 \).
The mass at \( A \) is \( 5 \).
Now consider the line segment \( AP \).
The point \( G \) is the intersection of cevians, which acts as the center of mass of the system.
\( G \) must balance the mass at \( A \) and the combined mass at \( P \).
Therefore, \( m_A \cdot AG = m_P \cdot GP \).
Substitute the known masses:
\( 5 \cdot AG = 7 \cdot GP \)
\( \frac{AG}{GP} = \frac{7}{5} \)
Thus, \( G \) divides \( AP \) internally in the ratio \( 7 : 5 \).
Step 4: Final Answer:
The required ratio is \( 7 : 5 \).