On dividing $p(x) = 2x^3 - 3x^2 + ax - 3a + 9$ by $(x + 1)$,if the remainder is $16$,then find the value of $a$. Then,find the remainder on dividing $p(x)$ by $(x + 2)$.

(A) According to the Remainder Theorem,if $p(x)$ is divided by $(x + 1)$,the remainder is $p(-1)$.
Given $p(-1) = 16$,we have:
$2(-1)^3 - 3(-1)^2 + a(-1) - 3a + 9 = 16$
$-2 - 3 - a - 3a + 9 = 16$
$4 - 4a = 16$
$-4a = 12$
$a = -3$.
Now,substitute $a = -3$ into $p(x)$:
$p(x) = 2x^3 - 3x^2 - 3x - 3(-3) + 9 = 2x^3 - 3x^2 - 3x + 18$.
To find the remainder when $p(x)$ is divided by $(x + 2)$,we calculate $p(-2)$:
$p(-2) = 2(-2)^3 - 3(-2)^2 - 3(-2) + 18$
$p(-2) = 2(-8) - 3(4) + 6 + 18$
$p(-2) = -16 - 12 + 6 + 18 = -4$.
Thus,$a = -3$ and the remainder is $-4$.