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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 $: \text{D}$