Option 3 : 124

**Calculation:**

Sum of roots = 2p

⇒ Here, 2p = 2r + 4r

⇒ p = 3r

Product of roots = 4q

⇒ 4q = 2r × 4r

⇒ 2q = 4r^{2}

Now, 3p^{2} + 2q

⇒ 3 × (3r)^{ 2} + 4r^{2}

⇒ (27 + 4) r^{2}

⇒ 31 r^{2}

For r = 2, 31 × r^{2} = 124

**∴ The required answer is 124**

__Key Points__

Sum of roots = -b/ a

Product of roots = c/ a

Subtraction of roots = √D/ a

where equation is ax^{2} + bx + c and D = √ (b^{2} - 4ac)