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

Find the shortest distance between the lines whose vector equations are $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})$ and $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$.

If $A(1,2,3), B(3,7,-2), C(6,7,7)$ and $D(-1,0,-1)$ are points in a plane, then the vector equation of the line passing through the centroids of $\triangle ABD$ and $\triangle ACD$ 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

If the line passing through the points $(a, 1, 6)$ and $(3, 4, b)$ crosses the $yz$-plane at the point $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$,then:

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}$ and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{3}}$,then the sum of possible values of $\lambda$ is