Mathematics - Vector Question with Solution | TestHub

Given equations are

$\frac{x-\lambda }{3}=\frac{y-2\sqrt{6}}{4}=\frac{z+2\sqrt{6}}{5}$ is passing through a point $\left(\lambda ,2\sqrt{6},3\sqrt{6}\right)$ and it's direction ratios are $3,4,5$.

So, 

${\overrightarrow{a}}_1=\lambda \hat{i}+2\sqrt{6}\hat{j}-2\sqrt{6}\hat{k}$

${\overrightarrow{b}}_1=3\hat{i}+4\hat{j}+5\hat{k}$

And,

$\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$ is passing through a point $\left(-\sqrt{6},\sqrt{6},\sqrt{6}\right)$ and it's direction ratios are $1,2,3$.

${\overrightarrow{a}}_2=-\sqrt{6}\hat{i}+\sqrt{6}\hat{j}+\sqrt{6}\hat{k}$

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

Now,

${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=\left(-\sqrt{6}\hat{i}+\sqrt{6}\hat{j}+\sqrt{6}\hat{k}\right)-\left(\lambda \hat{i}+2\sqrt{6}\hat{j}-2\sqrt{6}\hat{k}\right)=-\left(\sqrt{6}+\lambda \right)\hat{i}-\sqrt{6}\hat{j}+3\sqrt{6}\hat{k}$

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

$\Rightarrow {\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left(\hat{i}-2\hat{j}+\hat{k}\right)$

So,

$\Rightarrow \left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|=\sqrt{6}$

We know that the shortest distance between two skew lines is $\frac{\left|\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right)·\left({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right)\right|}{\left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|}=6$

$\Rightarrow \left|\frac{\left(\left(-\sqrt{6}-\lambda \right)\hat{i}-\sqrt{6}\hat{j}+3\sqrt{6}\hat{k}\right)·\left(\hat{i}-2\hat{j}+\hat{k}\right)}{\sqrt{6}}\right|=6$

$\Rightarrow \left|\frac{4\sqrt{6}-\lambda }{\sqrt{6}}\right|=6$

$\Rightarrow \lambda =10\sqrt{6}, -2\sqrt{6}$

So, required sum is

$={\left(10\sqrt{6}-2\sqrt{6}\right)}^2=384$