Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines $x + (k-1)y + 3 = 0$ and $2x + k^2 y - 4 = 0$. If the line $x - y + 2 = 0$ intersects the circle at the points A and B, then $(AB)^2$ is equal to:

The given lines are $x + (k-1)y + 3 = 0$ and $2x + k^2 y - 4 = 0$.

Since they are mutually perpendicular, the product of their slopes is $-1$:

$\left(\dfrac{-1}{k-1}\right) \left(\dfrac{-2}{k^2}\right) = -1$

$\dfrac{2}{k^2(k-1)} = -1 \Rightarrow k^3 - k^2 + 2 = 0$

By trial, $k = -1$ is a root. Factoring gives $(k+1)(k^2 - 2k + 2) = 0$. Since $k^2 - 2k + 2 = 0$ has no real roots, $k = -1$.

Substituting $k = -1$ into the equations of the lines, we get:

$x - 2y + 3 = 0$

$2x + y - 4 = 0$

Solving these two equations, we get the point of intersection as $x = 1$ and $y = 2$. Thus, the centre of the circle is $C(1, 2)$.

Since the circle passes through the origin $(0,0)$, the square of its radius $R$ is:

$R^2 = (1 - 0)^2 + (2 - 0)^2 = 5$

The perpendicular distance $d$ from the centre $C(1, 2)$ to the line $x - y + 2 = 0$ is:

$d = \dfrac{|1 - 2 + 2|}{\sqrt{1^2 + (-1)^2}} = \dfrac{1}{\sqrt{2}}$

The length of the chord $AB$ is given by $2\sqrt{R^2 - d^2}$. Therefore, its square is:

$(AB)^2 = 4(R^2 - d^2) = 4\left(5 - \dfrac{1}{2}\right) = 4 \times \dfrac{9}{2} = 18$

Answer: $18$