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 $x^2 + y^2 + 2gx + 2fy + 25 = 0$.

The centre of the circle is $(-g, -f)$. Since it lies in the first quadrant, $-g > 0$ and $-f > 0$, which implies $g < 0$ and $f < 0$.

The centre lies on the line $2x - y = 4$, so:

$-2g - (-f) = 4 \Rightarrow f = 2g + 4$

Let the radius of the circle be $R$. The side length $a$ of an equilateral triangle inscribed in the circle is $a = \sqrt{3}R$.

The area of the equilateral triangle is given as $27\sqrt{3}$:

$\dfrac{\sqrt{3}}{4}a^2 = 27\sqrt{3} \Rightarrow \dfrac{\sqrt{3}}{4}(3R^2) = 27\sqrt{3} \Rightarrow R^2 = 36$

The radius of the circle is also given by $R^2 = g^2 + f^2 - c$. Here $c = 25$, so:

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

Substituting $f = 2g + 4$ into the equation:

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

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

$(5g - 9)(g + 5) = 0$

Since $g < 0$, we get $g = -5$.

Then, $f = 2(-5) + 4 = -6$.

The centre of the circle is $(5, 6)$ and its radius is $R = 6$.

The distance $d$ from the centre $(5, 6)$ to the line $x = 1$ is:

$d = |5 - 1| = 4$

The length of the chord $L$ on the line $x = 1$ is given by:

$L = 2\sqrt{R^2 - d^2} = 2\sqrt{36 - 16} = 2\sqrt{20}$

The square of the length of the chord is:

$L^2 = (2\sqrt{20})^2 = 4 \times 20 = 80$

Answer: $80$