Let R₁ and R₂ are remainders when the polynomial x³ + 2x² - 5ax - 7 and x³ + ax² - 12x + 6 are divided by (x + 1) and (x - 2) respectively. If 2R₁ + R₂ = 6; find the value of a.

On dividing x³ + 2x² - 5ax - 7 by (x + 1), we get remainder R₁ :

∴ R₁ = (-1)³ + 2(-1)² - 5a(-1) - 7

⇒ R₁ = -1 + 2 + 5a - 7

⇒ R₁ = 5a - 6.

On dividing x³ + ax² - 12x + 6 by (x - 2), we get remainder R₂

∴ R₂ = (2)³ + a(2)² - 12(2) + 6

⇒ R₂ = 8 + 4a - 24 + 6

⇒ R₂ = 4a - 10.

Given,

⇒ 2R₁ + R₂ = 6

⇒ 2(5a - 6) + 4a - 10 = 6

⇒ 10a - 12 + 4a - 10 = 6

⇒ 14a - 22 = 6

⇒ 14a = 28

⇒ a = $\dfrac{28}{14}$ = 2.

Hence, a = 2.