The distance of the point $A(7, -2, 11)$ from the line $\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$ measured along the line $\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6}$ is:

The shortest distance between the skew lines $r=(6 \hat{i}+2 \hat{j}+2 \hat{k})+t(\hat{i}-2 \hat{j}+2 \hat{k})$ and $r=(-4 \hat{i}-\hat{k})+s(3 \hat{i}-2 \hat{j}-2 \hat{k})$ is

If the Cartesian equation of a line is $6x-2=3y+1=2z-2$,then the vector equation of the line is

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 all possible values of $\lambda$ is

The equation of the line passing through $(2, 3, 4)$ and parallel to the $Y$-axis is . . . . . . .

The distance of the point $P(-3, 2, 3)$ from the line passing through $A(4, 6, -2)$ with direction ratios $\langle -1, 2, 3 \rangle$ is . . . . . . units.