Explanation

(a)
Since, $\alpha,\ \beta,\ \gamma$ are the roots of the equation
$2x^{3} - 3x^{2} + 6x + 1 = 0$
Here, $\alpha + \beta + \gamma = \frac{3}{2}$ ...(i)
$\alpha\beta + \beta\gamma + \gamma\alpha = 3$ ...(ii)
And $\alpha\beta\gamma = - \frac{1}{2}$ ...(iii)
On squaring Eq. (i), we get
$\alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha\beta + \beta\gamma + \gamma\alpha) = \frac{9}{4}$
$\Rightarrow \ \ \ \ \alpha^{2} + \beta^{2} + \gamma^{2} = \frac{9}{4} - 2(3) = - \frac{15}{4}$ [from Eq. (ii)]