Let $z = x + iy$ be a complex number where $x$ and $y$ are integers. Then the area of the rectangle whose vertices are the roots of the equation: $z\bar{z}^{3} + \bar{z}z^{3} = 350$ is

$z = x + iy, \quad x, y \in \mathbb{Z}$

$z\bar{z}^{3} + \bar{z}z^{3} = 350$

$\text{To find: the area of the rectangle whose vertices are the roots}$

$z = re^{i\theta}, \quad \bar{z} = re^{-i\theta}$

$\bar{z}^{3} = r^{3}e^{-3i\theta}, \quad z^{3} = r^{3}e^{3i\theta}$

$z\bar{z}^{3} = r^{4}e^{-2i\theta}, \quad \bar{z}z^{3} = r^{4}e^{2i\theta}$

$z\bar{z}^{3} + \bar{z}z^{3} = r^{4}\left(e^{2i\theta} + e^{-2i\theta}\right) = 2r^{4}\cos 2\theta$

$2r^{4}\cos 2\theta = 350 \ \Rightarrow\ r^{4}\cos 2\theta = 175$

$r^{2} = x^{2} + y^{2}$

$r^{2}\cos 2\theta = x^{2} - y^{2}$

$r^{4}\cos 2\theta = (x^{2} + y^{2})(x^{2} - y^{2}) = 175$

$x^{4} - y^{4} = 175$

$(x^{2} - y^{2})(x^{2} + y^{2}) = 175$

$a = x^{2} - y^{2}, \quad b = x^{2} + y^{2}$

$ab = 175, \quad b > a > 0$

$x^{2} = \frac{a + b}{2}, \quad y^{2} = \frac{b - a}{2}$

$175 = 1 \times 175 = 5 \times 35 = 7 \times 25$

$(1,175):\ x^{2} = \frac{176}{2} = 88\ \text{✗}$

$(5,35):\ x^{2} = \frac{40}{2} = 20\ \text{✗}$

$(7,25):\ x^{2} = \frac{32}{2} = 16\ \checkmark$

$x^{2} = 16 \ \Rightarrow\ x = \pm 4$

$y^{2} = \frac{25 - 7}{2} = 9 \ \Rightarrow\ y = \pm 3$

$z = \pm 4 \pm 3i$

$\text{width} = 4 - (-4) = 8, \quad \text{height} = 3 - (-3) = 6$

$\text{Area} = 8 \times 6 = 48$

$\boxed{\text{Area} = 48} \quad \Rightarrow \text{option (1)}$