Let the centre of the circle $$x^2 + y^2 + 2gx + 2fy + 25 = 0$$ be in the first quadrant and lie on the line $$2x - y = 4$$. Let the area of an equilateral triangle inscribed in the circle be $$27\sqrt{3}$$. Then the square of the length of the chord of the circle on the line $$x = 1$$ is _______.

The equation of the circle is $$x^2 + y^2 + 2gx + 2fy + 25 = 0$$.

Center is $$C(-g, -f)$$.

Since the center lies in the first quadrant: $$-g > 0 \implies g < 0$$ and $$-f > 0 \implies f < 0$$.

The center lies on $$2x - y = 4 \implies 2(-g) - (-f) = 4 \implies f = 2g + 4$$.

Radius $$R = \sqrt{g^2 + f^2 - 25}$$.

$$\text{Area} = \frac{3\sqrt{3}}{4}R^2$$

$$\frac{3\sqrt{3}}{4}R^2 = 27\sqrt{3} \implies R^2 = 36 \implies R = 6$$

$$g^2 + f^2 - 25 = 36 \implies g^2 + f^2 = 61$$

$$g^2 + (2g + 4)^2 = 61$$

$$5g^2 + 16g - 45 = 0$$

$$(5g - 9)(g + 5) = 0 \implies g = -5 \quad (\text{since } g < 0)$$

$$f = 2(-5) + 4 = -6$$

Thus, the center of the circle is $$C(-g, -f) = (5, 6)$$

The perpendicular distance ($$d$$) from the center $$(5, 6)$$ to the vertical line $$x = 1$$ is: $$d = |5 - 1| = 4$$

$$L = 2\sqrt{R^2 - d^2}$$

$$L^2 = 4(R^2 - d^2) = 4(36 - 16) = 4(20) = 80$$