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

The shortest distance between lines $L_1$ and $L_2$,where $L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and $L_2$ is the line passing through the points $A(-4,4,3)$ and $B(-1,6,3)$ and perpendicular to the line $\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$,is

The lines $\frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 3}{0}$ and $\frac{x - 2}{0} = \frac{y - 3}{0} = \frac{z - 4}{1}$ are:

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

If the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular,find the value of $k$.

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}$ is