JEE Main 2023 — Three Dimensional Geometry Question with Solution

Equation of line joining $\left(1,2,3\right)$ and $\left(2,3,4\right)$ is

$\overrightarrow{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(\hat{i}+\hat{j}+\hat{k}\right)$

Also, vector equation is

$\overrightarrow{r}=\left(\hat{i}-\hat{j}+2\hat{k}\right)+\mu \left(2\hat{i}-\hat{j}\right)$

So,

${\overrightarrow{a}}_1=\left(\hat{i}+2\hat{j}+3\hat{k}\right)$

${\overrightarrow{a}}_2=\left(\hat{i}-\hat{j}+2\hat{k}\right)$

Hence,

${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=0\hat{i}-3\hat{j}-\hat{k}$

And,

${\overrightarrow{b}}_1=\hat{i}+\hat{j}+\hat{k}$

${\overrightarrow{b}}_2=2\hat{i}-\hat{j}+0\hat{k}$

And, 

${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& -1& 0\end{array}\right|=\hat{i}+2\hat{j}-3\hat{k}$

$\Rightarrow \left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|=\sqrt{14}$

Required shortest distance is

$\alpha =\frac{\left|\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right)·\left({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right)\right|}{\left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|}$

$\Rightarrow \alpha =\left|\frac{\left(-3\hat{j}-\hat{k}\right)·\left(\hat{i}+2\hat{j}-3\hat{k}\right)}{\sqrt{14}}\right|$

$\Rightarrow \alpha =\left|\frac{-6+3}{\sqrt{14}}\right|=\frac{3}{\sqrt{14}}$

Now, $28{\alpha }^2=228\times \frac{9}{14}=18$.