If the shortest distance between the lines $\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is $1$,then the sum of all possible values of $\lambda$ is:

Find the position vector of the image of the point with position vector $\vec{P} = 2\hat{i} + \hat{j} + 3\hat{k}$ in the line whose vector equation is $\vec{r} = \hat{j} - 2\hat{k} + \lambda(\hat{i} + \hat{j} - \hat{k})$.

Distance between the parallel lines $\frac{x}{3}=\frac{y-1}{-2}=\frac{z}{1}$ and $\frac{x+4}{3}=\frac{y-3}{-2}=\frac{z+2}{1}$ is

The shortest distance between the lines $\frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $\frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$ is:

If the shortest distance between the lines $r=(3 \hat{i}+4 \hat{j}-2 \hat{k})+t(-\hat{i}+2 \hat{j}+\hat{k})$ and $r=(\hat{i}-7 \hat{j}-2 \hat{k})+s(\hat{i}+3 \hat{j}+2 \hat{k})$ is equivalent to the projection of $P=-2 \hat{i}+11 \hat{j}$ on $Q$, then a possible vector $Q$ is

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is