On which of the following lines does the point of intersection of the line $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane $x+y+z=2$ lie?

  • A
    $\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+4}{-5}$
  • B
    $\frac{x-4}{1}=\frac{y-5}{1}=\frac{z-5}{-1}$
  • C
    $\frac{x-2}{2}=\frac{y-3}{2}=\frac{z+3}{3}$
  • D
    $\frac{x+3}{3}=\frac{4-y}{3}=\frac{z+1}{-2}$

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