Find the equation of the line which bisects the line segment joining points $\displaystyle \mathrm{A}(2,3,4)$ and $\displaystyle \mathrm{B}(4,5,8)$ and is perpendicular to the lines $\displaystyle \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ and $\displaystyle \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$.

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}=0and $\displaystyle 3 \mathrm{a}+8 \mathrm{b}-5 \mathrm{c}=0$\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} $$