JEE Main 2022 — Three Dimensional Geometry Question with Solution

We know that, shortest distance between two line is given by $\frac{\left|\left(a_2-a_1\right)·\left(b_1\times b_2\right)\right|}{\left|b_1\times b_2\right|}$

Here, $a_1=\left(1,2,3\right)$ $a_2=\left(2,4,5\right)$

${\overrightarrow{b}}_2=2\hat{i}+3\hat{j}+\lambda \hat{k}$ and ${\overrightarrow{b}}_2=\hat{i}+4\hat{j}+5\hat{k}$

So, S.D. $=\frac{\left|\left(\left(2-1\right)\hat{i}+\left(4-2\right)\hat{j}+\left(5-3\right)\hat{k}\right)·\left({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right)\right|}{\left|b_1\times b_2\right|}$

Calculating ${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& \lambda \\ 1& 4& 5\end{array}\right|$

$=\hat{i}\left(15-4\lambda \right)+\hat{j}\left(\lambda -10\right)+\hat{k}\left(5\right)$

$=\left(15-4\lambda \right)\hat{i}+\left(\lambda -10\right)\hat{j}+5\hat{k}$

So, $\left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|=\sqrt{{\left(15-4\lambda \right)}^2+{\left(\lambda -10\right)}^2+25}$

Now again, 

S.D. $=\frac{\left|\left(\hat{i}+2\hat{j}+2\hat{k}\right)·\left[\left(15-4\lambda \right)\hat{i}+\left(\lambda -10\right)\hat{j}+5k\right]\right|}{\sqrt{{\left(15-4\lambda \right)}^2+{\left(\lambda -10\right)}^2+25}}$

$\frac{\left|15-4\lambda +2\lambda -20+10\right|}{\sqrt{{\left(15-4\lambda \right)}^2+{\left(\lambda -10\right)}^2+25}}=\frac{1}{\sqrt{3}}$

squaring both side we get,

$3{\left(5-2\lambda \right)}^2=225+16{\lambda }^2-120\lambda +{\lambda }^2+100-20\lambda +25$

$12{\lambda }^2+75-60\lambda =17{\lambda }^2-140\lambda +350$

$5{\lambda }^2-80\lambda +275=0$

${\lambda }^2-16\lambda +55=0$

$\left(\lambda -5\right)\left(\lambda -11\right)=0$

$\Rightarrow \lambda =5,11$

So sum of $5+11=16$