If the equation of the plane passing through the point (1,1,2) and perpendicular to the line x-3 y+ 2 z-1=0=4 x-y+z is A x+ B y+ C z=1, then 140( C -…

If the equation of the plane passing through the point $\displaystyle (1,1,2)$ and perpendicular to the line $\displaystyle x-3 y+$ $\displaystyle 2 z-1=0=4 x-y+z$ is $\displaystyle \mathrm{A} x+\mathrm{B} y+\mathrm{C} z=1$, then $\displaystyle 140(\mathrm{C}-\mathrm{B}+\mathrm{A})$ is equal to $\displaystyle \_\_\_\_$.