Let the line L: (x-1)/2=(y+1)/-1=(z-3)/1 intersect the plane 2 x+y+3 z=16 at the point P. Let the point Q be the foot of perpendicular from the point…

Let the line $\displaystyle L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ intersect the plane $\displaystyle 2 x+y+3 z=16$ at the point $\displaystyle P$. Let the point $\displaystyle Q$ be the foot of perpendicular from the point $\displaystyle R(1,-1,-3)$ on the line $\displaystyle L$. If $\displaystyle \alpha$ is the area of triangle $\displaystyle P Q R$, then $\displaystyle \alpha^2$ is equal to $\displaystyle \_\_\_\_$.