# The shortest distance between the lines $\large\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\large\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is

$\begin{array}{1 1}(1)\large\frac{2}{\sqrt{3}}&(2)\frac{1}{\sqrt{6}}\\(3)\frac{2}{3}&(4)\frac{1}{2\sqrt{6}}\end{array}$

The equations of the given straight lies are
$\large\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$------(1)
$\large\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$-------(2)
The vector equations of the given straight lines are
$\overrightarrow{r}=(\overrightarrow{i}+\overrightarrow{2j}+\overrightarrow{3k})+t(\overrightarrow{2i}+3\overrightarrow{j}+4\overrightarrow{k})$------(3)
$\overrightarrow{r}=(2\overrightarrow{i}+4\overrightarrow{j}+\overrightarrow{5k})+s(-\overrightarrow{3i}+4\overrightarrow{j}+5\overrightarrow{k})$ -------(4)
Comparing equations (3) and (4)
$\overrightarrow{a_1}=\overrightarrow{i}+\overrightarrow{2j}+\overrightarrow{3k}$
$\overrightarrow{a_2}=2\overrightarrow{i}+4\overrightarrow{j}+\overrightarrow{5k}$
$\overrightarrow{u}=2\overrightarrow{i}+\overrightarrow{3j}+\overrightarrow{4k}$
$\overrightarrow{v}=3\overrightarrow{i}+\overrightarrow{4j}+\overrightarrow{5k}$
$\overrightarrow{a_2}-\overrightarrow{a_1}=\overrightarrow{i}+2\overrightarrow{j}+2 \overrightarrow{k}$
$[\overrightarrow{a_2} -\overrightarrow {a_1} \quad \overrightarrow{u} \quad \overrightarrow{v} ] = \begin{vmatrix} 1 & 2 & 2 \\ 2 & 3 & 4 \\3 & 4 &5 \end{vmatrix}$
$\qquad=1(15-16)-2(10-12)+2(8-9)$
$\qquad=-1+4-2=1$
$\overrightarrow{u} \times \overrightarrow{v}= \begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ 2 & 3 & 4 \\3 & 4 &5 \end{vmatrix}$
$\qquad= \overrightarrow{-i}+2 \overrightarrow{j}- \overrightarrow{k}$
$|\overrightarrow{u} \times \overrightarrow{v}|=|-\overrightarrow{i}+2 \overrightarrow{j}-\overrightarrow{k}|$
$\qquad= \sqrt {(-1)^2+(2)^2+(-1)^2}$
$\qquad= \sqrt {1+4+1}=\sqrt {6}$
Shortest distance $d= \large\frac{1}{\sqrt 6}$
Hence 2 is the correct answer.