Let the line $x - y = 4$ intersect the circle $C: (x-4)^2 + (y+3)^2 = 9$ at the points $Q$ and $R$. If $P(\alpha, \beta)$ is a point on $C$ such that $PQ = PR$, then $(6\alpha + 8\beta)^2$ is equal to __________.

The given circle $C: (x-4)^2 + (y+3)^2 = 9$ has its center at $C_0(4, -3)$ and radius $r = 3$.

The line $QR$ has the equation $x - y = 4$, which has a slope of $1$.

Since $PQ = PR$, the point $P(\alpha, \beta)$ must lie on the perpendicular bisector of the chord $QR$. The perpendicular bisector of any chord of a circle passes through its center. Therefore, the perpendicular bisector passes through $C_0(4, -3)$ and has a slope of $-1$ (since it is perpendicular to $QR$).

The equation of the perpendicular bisector is:
$y - (-3) = -1(x - 4)$
$x + y = 1$

Since $P(\alpha, \beta)$ lies on this perpendicular bisector, we have:
$\alpha + \beta = 1 \Rightarrow \beta = 1 - \alpha$

Also, $P(\alpha, \beta)$ lies on the circle $C$, so its coordinates must satisfy the circle's equation:
$(\alpha - 4)^2 + (\beta + 3)^2 = 9$

Substituting $\beta = 1 - \alpha$ into the equation:
$(\alpha - 4)^2 + (1 - \alpha + 3)^2 = 9$
$(\alpha - 4)^2 + (4 - \alpha)^2 = 9$
$2(\alpha - 4)^2 = 9$
$\alpha - 4 = \pm \dfrac{3}{\sqrt{2}}$
$\alpha = 4 \pm \dfrac{3}{\sqrt{2}}$

We need to find the value of $(6\alpha + 8\beta)^2$. Substituting $\beta = 1 - \alpha$:
$6\alpha + 8\beta = 6\alpha + 8(1 - \alpha) = 8 - 2\alpha$

Now, substituting the value of $\alpha$:
$8 - 2\alpha = 8 - 2\left(4 \pm \dfrac{3}{\sqrt{2}}\right) = \mp \dfrac{6}{\sqrt{2}} = \mp 3\sqrt{2}$

Squaring this value gives:
$(6\alpha + 8\beta)^2 = (\mp 3\sqrt{2})^2 = 18$

Answer: $18$