Let the plane P contain the line 2 x+y-z-3=0=5 x-3 y+4 z+9 and be parallel to the line (x+2)/2=(3-y)/-4=(z-7)/5. Then the distance of the point A(8,…

Let the plane P contain the line $\displaystyle 2 x+y-z-3=0=5 x-3 y+4 z+9$ and be parallel to the line $\displaystyle \frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point A($\displaystyle 8$, -$\displaystyle 1$, -$\displaystyle 19$) from the plane P measured parallel to the line $\displaystyle \frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$ is equal to $\displaystyle \_\_\_\_$.