Let S = \ z ∈ ℂ - i, 2i : z ^ 2 + 8iz - 15 z ^ 2 - 3iz - 2 ∈ ℝ\ . α...

Put $z = x + iy$ in the given expression and equate the imaginary part to zero.

$⇒ \dfrac{{z}^{2} + 8iz - 15}{{z}^{2} - 3iz - 2} = \dfrac{{\left(x + iy\right)}^{2} + 8i\left(x + iy\right) - 15}{{\left(x + iy\right)}^{2} - 3i\left(x + iy\right) - 2}$

$= \dfrac{{x}^{2} - {y}^{2} - 8y - 15 + i\left(2xy + 8x\right)}{{x}^{2} - {y}^{2} + 3y - 2 + i\left(2xy - 3x\right)}$

$= \dfrac{{x}^{2} - {y}^{2} - 8y - 15 + i\left(2xy + 8x\right)}{{x}^{2} - {y}^{2} + 3y - 2 + i\left(2xy - 3x\right)} \times \dfrac{\left({x}^{2} - {y}^{2} + 3y - 2 - i\left(2xy - 3x\right)\right)}{\left({x}^{2} - {y}^{2} + 3y - 2 - i\left(2xy - 3x\right)\right)}$

$= \dfrac{\left({x}^{2} - {y}^{2} - 8y - 15\right)\left({x}^{2} - {y}^{2} + 3y - 2\right) + \left(2xy + 8x\right)\left(2xy - 3x\right) + i\left(2xy + 8x\right)\left({x}^{2} - {y}^{2} + 3y - 2\right) - i\left(2xy - 3x\right)\left({x}^{2} - {y}^{2} - 8y - 15\right)}{{\left({x}^{2} - {y}^{2} + 3y - 2\right)}^{2} - {\left(2xy - 3x\right)}^{2}}$

But $Im\left(\dfrac{{z}^{2} + 8iz - 15}{{z}^{2} - 3iz - 2}\right) = 0$,

$⇒ - \left({x}^{2} - {y}^{2} - 8y - 15\right)\left(2xy - 3x\right) + \left(2xy + 8x\right)\left({x}^{2} - {y}^{2} + 3y - 2\right) = 0$

$⇒ \left({x}^{2} - {y}^{2}\right)\left(2xy + 8x - 2xy + 3x\right) + \left(8y + 15\right)\left(2xy - 3x\right) + \left(2xy + 8x\right)\left(3y - 2\right) = 0$

$⇒ 11{x}^{3} - 11x{y}^{2} + 16x{y}^{2} - 24xy + 30xy - 45x + 6x{y}^{2} - 4xy + 24xy - 16x = 0$

$⇒ 11{x}^{3} + 11x{y}^{2} + 26xy - 61x = 0$

$⇒ 11{x}^{2} + 11{y}^{2} + 26y - 61 = 0$

$∵ \alpha \ne 0$

So, put $y = - \dfrac{13}{11}$, $x = \alpha$
$11{\alpha}^{2} + 11 \cdot \dfrac{{13}^{2}}{{11}^{2}} - 26 \cdot \dfrac{13}{11} - 61 = 0$
$⇒ 121{\alpha}^{2} = 840$

$⇒ 242{\alpha}^{2} = 1680$

Hence this is the required answer.