On dividing $p(x) = x^{3} + 2x^{2} - 5ax - 7$ by $(x + 1)$,the remainder is $R_{1}$ and on dividing $q(x) = x^{3} + ax^{2} - 12x + 6$ by $(x - 2)$,the remainder is $R_{2}$. If $2R_{1} + R_{2} = 6$,then find the value of $a$.

(A) According to the Remainder Theorem,if a polynomial $f(x)$ is divided by $(x - c)$,the remainder is $f(c)$.
For $p(x) = x^{3} + 2x^{2} - 5ax - 7$ divided by $(x + 1)$,$R_{1} = p(-1) = (-1)^{3} + 2(-1)^{2} - 5a(-1) - 7 = -1 + 2 + 5a - 7 = 5a - 6$.
For $q(x) = x^{3} + ax^{2} - 12x + 6$ divided by $(x - 2)$,$R_{2} = q(2) = (2)^{3} + a(2)^{2} - 12(2) + 6 = 8 + 4a - 24 + 6 = 4a - 10$.
Given $2R_{1} + R_{2} = 6$,substitute the expressions for $R_{1}$ and $R_{2}$:
$2(5a - 6) + (4a - 10) = 6$
$10a - 12 + 4a - 10 = 6$
$14a - 22 = 6$
$14a = 28$
$a = 2$.