(A) Let $p(x) = x^{3}-3x^{2}+ax+24$. Since $(x-2)$ is a factor of $p(x)$,by the Factor Theorem,$p(2) = 0$. Substituting $x = 2$ into the polynomial: $

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(A) Let $p(x) = x^{3}-3x^{2}+ax+24$. Since $(x-2)$ is a factor of $p(x)$,by the Factor Theorem,$p(2) = 0$. Substituting $x = 2$ into the polynomial: $p(2) = (2)^{3} - 3(2)^{2} + a(2) + 24 = 0$. $8 - 3(4) + 2a + 24 = 0$. $8 - 12 + 2a + 24 = 0$. $20 + 2a = 0$. $2a = -20$. $a = -10$.