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