Question

The perpendicular bisector of a line segment with end points $\left(1,2,6\right)$ and $\left(-3,6,2\right)$ passes through $\left(-6,2,4\right)$ and has the equation of the form $\frac{x+6}{l}=\frac{y-2}{m}=\frac{z-4}{n}$ (where $l,m,n$ are integers, $l$ is a prime number and $l>0$ ), then the value of $lmn-\left(l+m+n\right)$ is equal to

• A. $-3$
• B. $-5$
• C. $-7$
• D. $-9$

Answer 1

Answer: (C)

Midpoint of the line segment is $\left(\frac{1-3}{2},\frac{2+6}{2},\frac{6+2}{2}\right)\equiv \left(-1,4,4\right)$
Parallel vector to the required line $=\left(-1+6\right)\hat{i}+\left(4-2\right)\hat{j}+\left(4-4\right)\hat{k}$
$=5\hat{i}+2\hat{j}+0\hat{k}$
Hence, the equation of the line is
$\frac{x+6}{5}-\frac{y-2}{2}=\frac{z-4}{0}$
$\Rightarrow l=5,m=2,n=0$
So, $lmn-\left(l+m+n\right)=0-7=-7$