Question

The perpendicular bisector of the line segment with endpoints $(2,3,2)$ and $(-4,1,4)$ passes through the point $(-3,6,1)$ and has equation of the form $\frac{x+3}{a}=\frac{y-6}{b}=\frac{z-1}{c}$ where $a,b$ and $c$ are relatively prime integers with $a>0$. The value of $abc-(a+b+c)$ is equal to

• A. $-1$
• B. $-2$
• C. $-3$
• D. $-4$

Answer 1

The correct option is B $-2$

The midpoint of the segment is $(\frac{2-4}{2},\frac{3+1}{2},\frac{2+4}{2})=(-1,2,3)$ and the vector between this point and $(-3,6,1)$ is $<-1-(-3),2-6,3-1>$ and $<2,-4,2>$.
$\therefore$ the line has parametric equations
$x=-3+2t,y=6-4t$ and $z=1+2r$.
Solving each of these equations for $t$ and setting the expressions equal makes the equation.
$\frac{x+3}{a}=\frac{y-6}{b}=\frac{z-1}{c}\Rightarrow \frac{x+3}{1}=\frac{y-6}{-2}$
$=\frac{z-6}{1}$, so $1\cdot (-2)\cdot 1-(1-2+1)=-2$.