Explanation

(a)
Given equation is $x^{2} - 2ax + a^{2} + a - 3 = 0.$
If roots are real then $D \geq 0$
$\Rightarrow 4a^{2} - 4(a^{2} + a - 3) \geq 0$
$\Rightarrow \ - a + 3 \geq 0$
$\Rightarrow \ a - 3 \leq 0 \Rightarrow a \leq 3$
As roots are less than 3, hence $f(3) > 0$
$9 - 6a + a^{2} + a - 3 > 0$
$\Rightarrow a^{2} - 5a + 6 > 0$
$\Rightarrow (a - 2)(a - 3) > 0$
$\Rightarrow Either\ a < 2\ or\ a > 3.$
Hence, only $a < 2$ satisfy.