Given equation of line is
$\frac{\mathit{x}-1}{2}$ = $\frac{\mathit{y}-2}{3}$ = $\frac{\mathit{z}-4}{6}$
This can also be written in the standard form as
$\frac{\mathit{x}-1}{2}$ = $\frac{\mathit{y}-2}{3}$ = $\frac{\mathit{z}-\left(-4\right)}{6}$
The vector form of the above equation is,
$\overrightarrow{\mathit{r}}$ = $\left(\hat{\mathit{i}}+2\hat{\mathit{j}}-4\mathit{k}\right)$ + $\mathit{\lambda }\left(2\hat{\mathit{i}}+3\hat{\mathit{j}}+6\hat{\mathit{k}}\right)$
⇒ $\overrightarrow{\mathit{r}}$ = ${\overrightarrow{\mathit{a}}}_1+\mathit{\lambda }\overrightarrow{\mathit{b}}$ ... (i)
where, $\overrightarrow{\mathit{a}}$ = $\hat{\mathit{i}}+2\hat{\mathit{j}}-4\hat{\mathit{k}}$ and $\overrightarrow{\mathit{b}}$ = $2\hat{\mathit{i}}+3\hat{\mathit{j}}+6\hat{\mathit{k}}$
The second equation of line is
$\frac{\mathit{x}-3}{4}$ = $\frac{\mathit{y}-3}{6}$ = $\frac{\mathit{z}+5}{12}$
The above equation can also be written as $\frac{\mathit{x}-3}{4}$ = $\frac{\mathit{y}-3}{6}$ = $\frac{\mathit{z}-\left(-5\right)}{12}$
The vector form of this equation is
$\overrightarrow{\mathit{r}}$ = $\left(3\hat{\mathit{i}}+3\hat{\mathit{j}}-5\mathit{k}\right)$ + $\mathit{\mu }\left(4\hat{\mathit{i}}+6\hat{\mathit{j}}+12\hat{\mathit{k}}\right)$
⇒ $\overrightarrow{\mathit{r}}$ = $\left(3\hat{\mathit{i}}+3\hat{\mathit{j}}-5\mathit{k}\right)$ + $2\mathit{\mu }\left(2\hat{\mathit{i}}+3\hat{\mathit{j}}+6\hat{\mathit{k}}\right)$
⇒ $\overrightarrow{\mathit{r}}$ = ${\overrightarrow{\mathit{a}}}_2+2\mathit{\mu }\overrightarrow{\mathit{b}}$ ... (ii)
where ${\overrightarrow{\mathit{a}}}_2$ = $\left(3\hat{\mathit{i}}+3\hat{\mathit{j}}-5\mathit{k}\right)$ and $\overrightarrow{\mathit{b}}$ = $2\hat{\mathit{i}}+3\hat{\mathit{j}}+6\hat{\mathit{k}}$
Since $\overrightarrow{\mathit{b}}$ is same in equations (1) and (2), the two lines are parallel. Distance d, between the two parallel lines is given by the formula,
d = $\left|\frac{\overrightarrow{\mathit{b}}\times (\overrightarrow{{\mathit{a}}_2}-\overrightarrow{{\mathit{a}}_1}}{\left|\mathit{b}\right|}\right|$
Here, $\overrightarrow{\mathit{b}}$ = $2\hat{\mathit{i}}+3\hat{\mathit{j}}+6\hat{\mathit{k}}$ , ${\overrightarrow{\mathit{a}}}_2$ = $\left(3\hat{\mathit{i}}+3\hat{\mathit{j}}-5\mathit{k}\right)$ and ${\overrightarrow{\mathit{a}}}_1$ = $\hat{\mathit{i}}+2\hat{\mathit{j}}-4\hat{\mathit{k}}$
On substitution, we get
d =
= $\frac{1}{\sqrt{49}}$
= $\frac{1}{7}|\begin{array}{ccc}\hat{\mathit{i}}& \hat{\mathit{j}}& \hat{\mathit{k}}\\ 2& 3& 6\\ 2& 1& -1\end{array}|$
= $\frac{1}{7}$ |$\hat{\mathit{i}}$ (- 3 - 6) - $\hat{\mathit{j}}$ (- 2 - 12) + $\hat{\mathit{k}}$ (2 - 6)|
= $\frac{1}{7}|-9\hat{\mathit{i}}+14\hat{\mathit{j}}-4\hat{\mathit{k}}|$
= $\frac{1}{7}|\sqrt{81+196+16}|$
= $\frac{\sqrt{293}}{7}$
Thus, the distance between the two given lines is $\frac{\sqrt{293}}{7}$