Question

Line $L$ meets lines $L_1:\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and $L_2:\frac{x-2}{2}=\frac{y-1}{4}=\frac{z-4}{5}$ orthogonally at points $P$ and Q. $(PQ{)}^2$ is D. DRs of line $L$ are $(a,b,c)\{a,b,c\in I\}$, then least value of $\frac{5D}{3}+|a|+|b|+|c|$.

Answer 1

Answer: 6

PQ distance is shortest distance between $L_1$ and $L_2$.
S.D. $=\frac{|[2\hat{i}+\hat{j}+4\hat{k}\hat{i}+2\hat{j}+3\hat{k}2\hat{i}+4\hat{j}+5\hat{k}]|}{|(\hat{i}+2\hat{j}+3\hat{k})\times (2\hat{i}+4\hat{j}+5\hat{k})|}$

$=\frac{3}{|-2\hat{i}+\hat{j}|}=\frac{3}{\sqrt{5}}$
or $(PQ{)}^2=\frac{9}{5}\Rightarrow 5D=9$
DRs of line $L$ are proportional to $(-2,1,0)$.