If the shortest distance between the lines x - 4 1 = y + 1 2 = z - 3 and x - λ 2 = y + 1 4 = z - 2 - 5 is 6…

$\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$

$\frac{x-\lambda }{2}=\frac{y+1}{4}=\frac{z-2}{-5}$

The shortest distance between the lines
$=\left|\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})·\left(\overrightarrow{{\mathrm{d}}_1}\times \overrightarrow{{\mathrm{d}}_2}\right)}{\left|\overrightarrow{{\mathrm{d}}_1}\times \overrightarrow{{\mathrm{d}}_2}\right|}\right|$

$a\equiv \left(4,-1,0\right), b\equiv \left(\lambda ,-1,2\right)$

$\Rightarrow \overrightarrow{a}-\overrightarrow{b}=\left(4\hat{i}-\hat{j}\right)-\left(\lambda \hat{i}-\hat{j}+2\hat{k}\right)$

$\Rightarrow \overrightarrow{a}-\overrightarrow{b}=\left(4-\lambda \right)\hat{i}-2\hat{k}$

So, the shortest distance is given by,

$D=\left|\frac{\left|\begin{array}{ccc}\lambda -4& 0& 2\\ 1& 2& -3\\ 2& 4& -5\end{array}\right|}{\left|\begin{array}{ccc}\hat{i}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 2& -3\\ 2& 4& -5\end{array}\right|}\right|$

$\Rightarrow D=\left|\frac{(\lambda -4)(-10+12)-0+2(4-4)}{|2\hat{i}-1\hat{j}+0\hat{k}|}\right|$

$\Rightarrow \frac{6}{\sqrt{5}}=\left|\frac{2(\lambda -4)}{\sqrt{5}}\right|$

$\Rightarrow 3=|\lambda -4|$

$\Rightarrow \lambda -4=\pm 3$

$\Rightarrow \lambda =7,1$

So, sum of all possible values of $\lambda$ is $8$.