Find the equation of the line which bisects the line segment joining points A (2,3,4) and B (4,5,8) and is perpendicular to the lines…

Let direction ratios of the required line be $\displaystyle \mathrm{a}, \mathrm{b}, \mathrm{c}$. ∵ the required line is perpendicular to both the given lines $$\therefore 3 \mathrm{a}-16 \mathrm{b}+7 \mathrm{c}=0 $$ and $\displaystyle 3 \mathrm{a}+8 \mathrm{b}-5 \mathrm{c}=0$ $$\begin{aligned} & \Rightarrow \frac{\mathrm{a}}{24}=\frac{\mathrm{b}}{36}=\frac{\mathrm{c}}{72} \\ & \Rightarrow \frac{\mathrm{a}}{2}=\frac{\mathrm{b}}{3}=\frac{\mathrm{c}}{6} \end{aligned} $$ The mid-point of the line-segment AB is $\displaystyle (3,4,6)$ Hence, the required equation of the line is $$\frac{x-3}{2}=\frac{y-4}{3}=\frac{z-6}{6} $$