Explanation

(a)
Let roots be $\alpha$ and $2\alpha$
$\therefore\ \ \alpha + 2\alpha = 3\alpha = - \frac{(3a - 1)}{(a^{2} - 5a + 3)}$
And $\alpha.2\alpha = 2\alpha^{2} = \frac{2}{(a^{2} - 5a + 3)}$
$\Rightarrow \ \frac{(3a - 1)^{2}}{9\left( a^{2} - 5a + 3 \right)^{2}} = \frac{1}{(a^{2} - 5a + 3)}$
$\Rightarrow (3a - 1)^{2} = 9(a^{2} - 5a + 3)$
$\Rightarrow 45a - 6a = 27 - 1\ \ \ \Rightarrow \ \ \ \ a = \frac{2}{3}$