A) \[(-3,2,1)\]
B) \[(3,2,1)\]
C) \[(1,2,-3)\]
D) \[(-1,2,3)\]
Correct Answer: A
Solution :
Since the Plane BISECTS the line joining the points, then the Plane must meet the line at the Midpoint of the line which is \[(\frac{1-3}{2},\frac{2+4}{2},\frac{5+3}{2})=(\frac{-2}{2},\frac{6}{2},\frac{8}{2})=(-1,3,4)\] (As the line is perpendicular to the plane) Now, the direction cosines of the plane are \[(-3-1,4-2,5-3)=(-4,2,2)\] So the Equation of the Plane must be \[=-4x+2y+2z=\lambda \] and now since the midpoint of the line is lying in the plane ,it must satisfy the plane \[\therefore -4(-1)+2(3)+2(2)=\lambda \Rightarrow \lambda =18\] Therefore, Equation of plane \[\Rightarrow -4x+2y+2z=18\] Now, out of the given option only one point \[(-3,2,1)\]is satisfying the Plane as follows \[\Rightarrow -4(-3)+2(2)+2(1)=18\] Therefore Correct Answer is A
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