If $p(x) = x^{4} - 2x^{3} + 3x^{2} - ax + 3a - 7$ is divided by $(x + 1)$,the remainder is $19$. Find the value of $a$. Also,find the remainder when $p(x)$ is divided by $(x + 2)$.

(A) According to the Remainder Theorem,if $p(x)$ is divided by $(x + 1)$,the remainder is $p(-1)$.
Given $p(-1) = 19$.
$p(-1) = (-1)^{4} - 2(-1)^{3} + 3(-1)^{2} - a(-1) + 3a - 7 = 19$
$1 + 2 + 3 + a + 3a - 7 = 19$
$4a - 1 = 19$
$4a = 20 \implies a = 5$.
Now,$p(x) = x^{4} - 2x^{3} + 3x^{2} - 5x + 3(5) - 7 = x^{4} - 2x^{3} + 3x^{2} - 5x + 8$.
To find the remainder when $p(x)$ is divided by $(x + 2)$,we calculate $p(-2)$.
$p(-2) = (-2)^{4} - 2(-2)^{3} + 3(-2)^{2} - 5(-2) + 8$
$p(-2) = 16 - 2(-8) + 3(4) + 10 + 8$
$p(-2) = 16 + 16 + 12 + 10 + 8 = 62$.
Thus,$a = 5$ and the remainder is $62$.