Perpendiculars are drawn from points on the line x+2/2 = y+1/1 = z/3 to the plane x+y+z=3. The feet of perpendiculars lie on the line is

Equation of the given line x+2/2=y+1/1=z/3=λ
So any point on line is x=2λ-2, y=λ-1, z=3λ
If it lies on the plane x+y+z=3, then
(2λ-2)+(λ-1)+(3λ)=3
6λ-3=3
6λ=6
λ=1
The point of intersection of the line and the plane has coordinates A(0,0,3) (1)
Now, from the equation of the line, we observe that (-2,-1,0) is a point on it.
Let (x,y,z) be the foot of the perpendicular from point (-2,-1,0) on the plane x+y+z=3.
λ= -[1(-2)+1(-1)+0×1]/√1²+1²+1² = 3/√3 = √3
Therefore, x=-2+2λ, y=-1+λ, z=3λ
x=-2+2√3, y=-1+√3, z=3√3
B(x,y,z) ≈ (1.464,-0.268,5.196)
From (1) and the coordinates of B, direction ratios of AB are (1.464, -0.268, 2.196) ≈ (1,-1,2) (2)
Hence, from (1) and (2), equation of the required line is x/1=y/-1=z-3/2 = k
x=k, y=-k, z=2k+3
When k=0, the coordinates are (0,0,3), which is the point A.
When k=1, the coordinates are (1,-1,5)
Hence, the required line is x/1=y/-1=z-3/2.
Let's check the options:
If x=2, y=-2, z=5, this satisfies the condition.
Option D, x2=y-1=z/5 is the equation of the line.