Input: A right-angled triangle \( ABC \) with sides \( AB = 5 \) and \( BC = 12 \).
Calculation of \( \triangle ABC \) Area:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 5 \times 12 = 30 \)
Let \( BP = p \) and \( BQ = q \).
Area of \( \triangle ABP = \frac{1}{2} \times AB \times BP = \frac{1}{2} \times 5 \times p = 2.5p \).
Area of \( \triangle ABQ = \frac{1}{2} \times AB \times BQ = \frac{1}{2} \times 5 \times q = 2.5q \).
Condition 1: The area of \( \triangle ABC \) is 1.5 times the area of \( \triangle ABP \).
\( 30 = 1.5 \times 2.5p \)
\( 30 = 3.75p \)
\( p = \frac{30}{3.75} = 8 \)
Condition 2: The areas of \( \triangle ABP, \triangle ABQ, \triangle ABC \) form an arithmetic progression (A.P.).
This implies: \( 2 \times \text{Area of } ABQ = \text{Area of } ABP + \text{Area of } ABC \)
\( 2 \times 2.5q = 2.5 \times 8 + 30 \)
\( 5q = 20 + 30 \)
\( 5q = 50 \)
\( q = 10 \)
Result: The length of \( PQ \) is the difference between \( BQ \) and \( BP \).
\( PQ = BQ - BP = q - p = 10 - 8 = \boxed{2} \)