Official solution
Since the plane bisects the line joining the points (1, 2, 3) and (-3, 4, 5) then the plane passes through the midpoint of the line which is :
\[\left( {{{1 - 3} \over 2},{{2 + 4} \over 2},{{5 + 3} \over 2}} \right)\] \[ \equiv \] \[\left( {{{ - 2} \over 2},{6 \over 2},{8 \over 2}} \right)\] \[ \equiv \] (-1, 3, 4).
As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are (- 3 - 1, 4 - 2, 5 - 3) = (- 4, 2, 2)
So the equation of the plane is : - 4x + 2y + 2z = \[\lambda \]
As plane passes through (- 1, 3, 4)
So, - 4(- 1) + 2(3) + 2(4) = \[\lambda \] \[ \Rightarrow \] \[\lambda \] = 18
Therefore, equation of plane is : - 4x + 2y + 2z = 18
Now, only (- 3, 2, 1) satiesfies the given plane as
- 4(- 3) + 2(2) + 2(1) = 18