The shortest distance between the lines $\frac{x-1}{2} =\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-5}{8}$ is equal to

$\sqrt{\frac{5}{29}}$

The correct answer is Option (3) → $\sqrt{\frac{5}{29}}$

Given lines

$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$

$\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-5}{8}$

A point on first line is $A(1,2,3)$ and its direction vector is

$\vec a=(2,3,4)$

A point on second line is $B(2,4,5)$ and its direction vector is

$\vec b=(4,6,8)=2(2,3,4)$

Hence the lines are parallel

Shortest distance between two parallel lines is

$\frac{|(\vec{AB}\times\vec a)|}{|\vec a|}$

$\vec{AB}=(1,2,2)$

$\vec{AB}\times\vec a= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 2 & 2\\ 2 & 3 & 4 \end{vmatrix} =(2,0,-1)$

$|\vec{AB}\times\vec a|=\sqrt{5}$

$|\vec a|=\sqrt{29}$

Shortest distance $=\frac{\sqrt{5}}{\sqrt{29}}=\sqrt{\frac{5}{29}}$

The shortest distance between the given lines is $\sqrt{\frac{5}{29}}$.