Which of the following is the point of intersection of the line $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane $x+y+z=2$?

Let $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}=r$.Any point on the line can be represented as:$x=2r+4,y=2r+5,z=r+3$ Substitute these coordinates into the plane equation $x+y+z=2$:
$$(2r+4)+(2r+5)+(r+3)=2$$
$$5r+12=2⟹5r=-10⟹r=-2$$
Now, find the point:$x=2(-2)+4=0$$y=2(-2)+5=1$$z=(-2)+3=1$ The point of intersection is $(0,1,1)$.