Omega = ((1−i)(9−5i))/((9+5i)(9-5i))

$$(1-i)z + (1+i)\omega = 3$$ $$(1-i)z = 3 - (1+i)\omega$$ $$(1-i)z = 3 - (1+i)\left(\frac{2}{53} - \frac{7}{53}i\right)$$ $$(1-i)z = 3 - \frac{1}{53}(1+i)(2-7i)$$ $$(1-i)z = 3 - \frac{1}{53}(2 - 7i + 2i - 7i^2)$$ $$(1-i)z = 3 - \frac{1}{53}(2 - 5i + 7)$$ $$(1-i)z = 3 - \frac{1}{53}(9 - 5i)$$ $$(1-i)z = \frac{3 \times 53}{53} - \frac{9 - 5i}{53}$$ $$(1-i)z = \frac{159 - (9 - 5i)}{53}$$ $$(1-i)z = \frac{159 - 9 + 5i}{53}$$ $$(1-i)z = \frac{150 + 5i}{53}$$

$$z = \frac{150 + 5i}{53(1-i)}$$ $$z = \frac{150 + 5i}{53-53i}$$

Multiply the numerator and denominator by the conjugate of the denominator, $53+53i$:

$$z = \frac{150 + 5i}{53-53i} \times \frac{53+53i}{53+53i}$$ $$z = \frac{(150+5i)(53+53i)}{53^2 - (53i)^2}$$ $$z = \frac{150 \times 53 + 150 \times 53i + 5i \times 53 + 5i \times 53i}{53^2 - 53^2 i^2}$$ $$z = \frac{7950 + 7950i + 265i + 265i^2}{2809 + 2809}$$ $$z = \frac{7950 + 8215i - 265}{5618}$$ $$z = \frac{7685 + 8215i}{5618}$$ $$z = \frac{7685}{5618} + \frac{8215}{5618}i$$

Both numerator and denominator are divisible by 53:
$7685 \div 53 = 145$
$8215 \div 53 = 155$
$5618 \div 53 = 106$

$$z = \frac{145}{106} + \frac{155}{106}i$$

So, $\boxed{z = \frac{145}{106} + \frac{155}{106}i}$.