Explanation:

- We are given a triangle ABC with DE parallel to BC.

- By the Basic Proportionality Theorem (or Thales's Theorem), if a line is parallel to one side of a triangle and intersects the other two sides, it divides those sides proportionally.

- Given, AD = a, DB = a + 4, so AB = a + (a + 4) = 2a + 4.

- Also, AE = 2a + 3, EC = 7a, so AC = (2a + 3) + 7a = 9a + 3.

- According to the theorem mentioned:

$$

\frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{a}{a + 4} = \frac{2a + 3}{7a}

$$

- Cross-multiplying gives: \(7a^2 = (a + 4)(2a + 3)\)

- This simplifies to: \(7a^2 = 2a^2 + 3a + 8a + 12\)

- Simplifying further: \(7a^2 = 2a^2 + 11a + 12\)

- Rearrange: \(5a^2 - 11a - 12 = 0\)

- Using the quadratic formula: \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where a = 5, b = -11, c = -12.

- Discriminant: \(b^2 - 4ac = (-11)^2 - 4(5)(-12) = 121 + 240 = 361\)

- Roots are: \(a = \frac{11 \pm 19}{10}\)

- So, \(a = 3\) or a negative value.

- Since \(a > 0\), \(a = 3\).

- Correct Answer: Option:3 - 3