3D geometry jee adv 13 ques doubt....pls help me out .....

Any point B on line is (2 – 2, – – 1, 3)
Point B lies on the plane for some 
 (2 – 2) + (– – 1) + 3 = 3
 4 = 6  3
2
   B 1, 5 , 9
2 2
     
 
The foot of the perpendicular from point (– 2, – 1, 0) on the plane is the point A (0, 1, 2)
 D.R. of AB = 1, 7 , 5
2 2
  
 
 
 (2, –7, 5)
Hence x y 1 z 2
2 7 5