Let the image of the point P (0,-5,0) in the line (x-1)/2=y/1=(z+1)/-2 be the point R and the image of the point Q (0, -1/2, 0) in the line…

Let the image of the point $\displaystyle \mathrm{P}(0,-5,0)$ in the line $\displaystyle \frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2}$ be the point R and the image of the point $\displaystyle \mathrm{Q}\left(0, \frac{-1}{2}, 0\right)$ in the line $\displaystyle \frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1}$ be the point S. Then the square of the area of the parallelogram PQRS is $\displaystyle \_\_\_\_$.