Explanation

(a)
Since, the roots of the equation $4x^{3} - 12x + 11x + k = 0$ are in AP which are $\alpha - a,\ \alpha,\ \alpha + a$.
$\therefore$ Sum of roots, $3\alpha = \frac{12}{4} = 3\ \Rightarrow \ \alpha = 1$
Since, $\alpha$ is a root, therefore it satisfies the given equation
$ie,\ \ \ \ \ 4x^{3} - 12x^{2} + 11x + k = 0$
$\therefore 4 - 12 + 11 + k = 0\ \ \ \ \Rightarrow \ \ \ k = - 3$