Solution:

Finding the Midpoint of the Line:

- The midpoint of the line joining two points P(x1, y1, z1) and Q(x2, y2, z2) is given by:
- Midpoint = ((x1 + x2)/2, (y1 + y2)/2, (z1 + z2)/2)
- Here, the two points are (2, 3, 4) and (6, 7, 8), so the midpoint is:
- Midpoint = ((2 + 6)/2, (3 + 7)/2, (4 + 8)/2) = (4, 5, 6)

Finding the Direction Vector of the Line:

- The direction vector of the line joining two points P(x1, y1, z1) and Q(x2, y2, z2) is given by:
- Direction vector = (x2 - x1, y2 - y1, z2 - z1)
- Here, the two points are (2, 3, 4) and (6, 7, 8), so the direction vector is:
- Direction vector = (6 - 2, 7 - 3, 8 - 4) = (4, 4, 4)

Finding the Normal Vector of the Plane:

- The plane which bisects a line joining two points will pass through the midpoint of the line and will be perpendicular to the line.
- Therefore, the normal vector of the plane will be the same as the direction vector of the line.
- Hence, the normal vector of the plane is (4, 4, 4)

Writing the Equation of the Plane:

- The equation of a plane passing through a point (x1, y1, z1) and having a normal vector (a, b, c) is given by:
- a(x - x1) + b(y - y1) + c(z - z1) = 0
- Here, the midpoint of the line is (4, 5, 6) and the normal vector of the plane is (4, 4, 4).
- Substituting these values in the equation of the plane, we get:
- 4(x - 4) + 4(y - 5) + 4(z - 6) = 0
- Simplifying, we get:
- x + y + z - 15 = 0

Hence, the equation of the plane which bisects the line joining (2, 3, 4) and (6, 7, 8) is x + y + z - 15 = 0, which is option 'A'.

If the plane bisects the line, then plane has to pass through middle point of line
So middle point is (4,5,6).
And line and plane are perpendicular to each other.
so drs of plane will be (6-2,7-3,8-4,) or(4,4,4). So eqn of plane with drs (4,4,4) and passing through (4,5,6) will be = 4(x-4) +4(y-5)+ 4(z-6)=0.
So, x-4+y-5+z-6=0.
So, x+y+z-15=o.