Question

Sol. [Key Concept: The foot of perpendicular from $P(\lambda,2,3)$ to the given line is $Q(1,\mu,2)$, fixing $\lambda,\mu$; the two resulting parallel lines are then $\frac{\sqrt{146}}{7}$ apart.]

$Q(1,\mu,2)$ satisfies the line L

$${\therefore\frac{1 - 4}{1} = \frac{\mu - 9}{2} = \frac{2 - 5}{1} }{\Rightarrow - 3 = \frac{\mu - 9}{2} = - 3\ \Rightarrow \mu = 3 }$$D.R. of $PQ:\lambda - 1,2 - \mu,1$

Since $PQ\bot L \Rightarrow (\lambda - 1).1 + 2(2 - \mu) + 1.1 = 0$

$$\Rightarrow \lambda - 2\mu + 4 = 0\ \therefore\lambda = 2 $$Now lines : $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$ &

$$\frac{x - 2}{2} = \frac{y - 3}{3} = \frac{z + 5}{6} $$i.e. $\overrightarrow{r} = (\widehat{i} + 2\widehat{j} - 4\widehat{k}) + t_{1}(2\widehat{i} + 3\widehat{j} + 6\widehat{k})$

$$\&\overrightarrow{r} = (2\widehat{i} + 3\widehat{j} - 5\widehat{k}) + t_{2}(2\widehat{i} + 3\widehat{j} + 6\widehat{k}) $$Distance

$${= \frac{|((\widehat{i} + 2\widehat{j} - 4\widehat{k}) - (2\widehat{i} + 3\widehat{j} - 5\widehat{k})) \times (2\widehat{i} + 3\widehat{j} + 6\widehat{k})|}{|2\widehat{i} + 3\widehat{j} + 6\widehat{k}|} }{= \frac{| - 9\widehat{i} + 8\widehat{j} - \widehat{k}|}{7} = \frac{\sqrt{146}}{7} }$$