Let the line $\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{4}$ lie in the plane $x-y+az+b=0$. The value of $b-a$ is:

Since the line lies in the plane:The direction of the line $(2,-3,4)$ must be perpendicular to the normal of the plane $(1,-1,a)$.Their dot product is zero:$2(1)+(-3)(-1)+4(a)=0$$2+3+4a=0⟹4a=-5⟹a=-5/4$.The point $(1,-2,3)$ on the line must lie on the plane:$1-(-2)+a(3)+b=0$$3+3a+b=0$ Substitute $a=-5/4$:$3+3(-5/4)+b=0⟹3-15/4+b=0⟹-3/4+b=0⟹b=3/4$.The value of $b-a=3/4-(-5/4)=8/4=2$.