Circle

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