Find the angle between the two lines $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and $\frac{x-1}{10}=\frac{y+3}{2}=\frac{z+4}{-11}$.

If the lines $\frac{x-k}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+3}{1}$ are co-planar,then the value of $k$ is $.....$

Statement $1:$ The shortest distance between the lines $\frac{x}{2} = \frac{y}{-1} = \frac{z}{2}$ and $\frac{x-1}{4} = \frac{y-1}{-2} = \frac{z-1}{4}$ is $\sqrt{2}$.
Statement $2:$ The shortest distance between two parallel lines is the perpendicular distance from any point on one of the lines to the other line.

The equation of a line passing through the point $(-3, 2, -4)$ and equally inclined to the axes is:

If the lines $\frac{1-x}{2}=\frac{7y+4}{2\lambda}=\frac{2z-5}{2}$ and $\frac{7-7x}{3\lambda}=\frac{y-1}{7}=\frac{6-z}{5}$ are at right angles,then the value of $\lambda$ 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.