Question

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

• A. $(-3,2,1)$
• B. $(3,2,1)$
• C. $(1,2,-3)$
• D. $(-1,2,3)$

Answer 1

The correct option is A $(-3,2,1)$

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
$\therefore -4(-1)+2\left(3\right)+2\left(2\right)=\lambda \Rightarrow \lambda =18$
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
$\Rightarrow -4(-3)+2\left(2\right)+2\left(1\right)=18$
Therefore Correct Answer is $A$