Given that, $x^{2}-(A-3)x-(A-2) = 0 \quad \longrightarrow (1)$
Let the roots of a quadratic equation be $\alpha$ and $\beta$ respectively.
Now,
- Sum of roots $ = \alpha+\beta = A-3$
- Product of roots $ = \alpha\beta = -(A-2) = 2-A$
Given that, $\alpha^{2} + \beta^{2} = 0$
$\Rightarrow (\alpha+\beta)^{2}-2\alpha\beta = 0$
$\Rightarrow (A-3)^{2}-2(2-A) = 0$
$\Rightarrow A^{2}+9-6A-4+2A = 0$
$\Rightarrow A^{2}-4A+5 = 0$
$\Rightarrow A = \frac{4\pm\sqrt{16-20}}{2}$
$\Rightarrow A = \frac{4\pm\sqrt{-4}}{2}$
$\Rightarrow A = \frac{4\pm2i}{2} \quad [\because \sqrt{-1} = i]$
$\Rightarrow A = \frac{4+2i}{2},\frac{4-2i}{2}$
$\Rightarrow \boxed{A = 2+i, 2-i}$
Correct Answer $: \text{D}$