Question

The sum of the intercepts of the plane which bisects the line segment joining $(0,1,2)$ and $(2,3,0)$ perpendicularly is

• A. $2$
• B. $4$
• C. $6$
• D. $12$

Answer 1

Answer: (A)

$2$

The dr's of the normal to the plane will be $(2,2,-2)$.
Hence, the plane is , $2x+2y-2z=d$
The midpoint of the line segment is $(1,2,1)$.
Hence, $d=4$.
Hence, the intercepts are $(\frac{1}{2},1,\frac{1}{2})$
The sum of intercepts is $2$.