If the lines $\frac{2x-4}{\lambda} = \frac{y-1}{2} = \frac{z-3}{1}$ and $\frac{x-1}{1} = \frac{3y-1}{\lambda} = \frac{z-2}{1}$ are perpendicular to each other,then $\lambda=$

The shortest distance between lines $\bar{r}=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and $\bar{r}=(2 \hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}-\hat{j}+\hat{k})$ is

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other,then $k = \dots$

The sum of all values of $ \alpha $, for which the shortest distance between the lines $ \frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha} $ and $ \frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha} $ is $ \sqrt{2} $, is

The equation of the line of shortest distance between the lines $\frac{x - 6}{3} = \frac{y - 7}{-1} = \frac{z - 4}{1}$ and $\frac{x}{-3} = \frac{y + 9}{2} = \frac{z - 2}{4}$ is:

If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$,then the largest possible value of $|\lambda|$ is equal to ..........