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 $a b c-(a+b+c)$ is equal to

• A. -1
• B. -2
• C. -3
• D. -4

Answer 1

Answer: (B)

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