A plane bisects the line segment joining the points (1,2,3) and (x,4,5) at right angles. Then this plane also passes through the point: (x,2,1), (3,2,1), (1,2,x), (x,2,3)

Since the Plane BISECTS the line joining the points , then the Plane must meet the line at the Midpoint of the line which is
Let A = (1,2,3) and B = (x,4,5)
Midpoint M = [ (1+x)/2 , (2+4)/2 , (3+5)/2 ] = [ (1+x)/2 , 3 , 4 ]
The plane passes through the midpoint M.
Let the coordinates of the point be (a,b,c).
Then the plane passes through the point (a,b,c) if and only if the coordinates satisfy the equation of the plane.
Let's consider the options:
(3,2,1) : Substituting in the equation of the plane, we get
(3,2,1) does not lie on the plane.
(1,2,x) : Substituting this in the equation of the plane.
(x,2,3) : Substituting this in the equation of the plane.
(x,2,1) : Substituting this in the equation of the plane.
The midpoint of (1,2,3) and (x,4,5) is ((1+x)/2, 3, 4).
Let the plane be ax+by+cz = d.
The vector joining (1,2,3) and (x,4,5) is (x-1, 2, 2).
The normal vector to the plane is perpendicular to this vector.
The equation of the plane passing through ((1+x)/2, 3, 4) is given by:
a(x - (1+x)/2) + b(y - 3) + c(z - 4) = 0
This plane passes through (3,2,1) if a(3 - (1+x)/2) + b(2-3) + c(1-4) = 0
This plane passes through (x,2,1) if a(x - (1+x)/2) + b(2-3) + c(1-4) = 0
The plane bisects the line segment at right angles. The midpoint of the line segment is ((1+x)/2, 3, 4).
The normal vector to the plane is proportional to (x-1, 2, 2).
The equation of the plane is a(x-((1+x)/2)) + b(y-3) + c(z-4) = 0
If the point (3,2,1) lies on the plane, then a(3 - (1+x)/2) + b(-1) + c(-3) = 0
If x=3, then a(3 - 2) - b - 3c = 0 => a = b+3c
If (3,2,1) lies on the plane, then (3,2,1) satisfies the equation of the plane.