If the shortest distance between the lines\ $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}$ and $

Question

If the shortest distance between the lines\ $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}$ and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{3}}$, then the sum of all possible values of $\lambda$ is\

Professor PC Thomas Classes · Answer

(a) : Shortest distance\ $=\left|\frac{\left|\begin{array}{ccc}2-1 & 4-2 & 5-3 \ 2 & 3 & \lambda \ 1 & 4 & 5\end{array}ight|}{\sqrt{(15-4 \lambda)^{2}+(\lambda-10)^{2}+(8-3)^{2}}}ight|$\ $\Rightarrow \frac{1}{\sqrt{3}}=\frac{1(15-4 \lambda)-2(10-\lambda)+2(8-3)}{\sqrt{(15-4 \lambda)^{2}+(\lambda-10)^{2}+5^{2}}}$\ $\Rightarrow \frac{1}{3}=\frac{(5-2 \lambda)^{2}}{17 \lambda^{2}-140 \lambda+350}$\ $\Rightarrow \quad 17 \lambda^{2}-140 \lambda+350=12 \lambda^{2}-60 \lambda+75$\ $\Rightarrow 5 \lambda^{2}-80 \lambda+275=0 \Rightarrow \lambda=5$ or 11\ $	herefore \quad$ Required sum $=16$\