Lines $\frac{2x-5}{k} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other. Then,the value of $k$ is . . . . . . .

If lines $\frac{2x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$ and $\frac{x-1}{1}=\frac{3y-1}{\lambda}=\frac{z-2}{1}$ are perpendicular to each other,then $\lambda = \ldots$.

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:

The line passing through the points $(a, 1, 6)$ and $(3, 4, b)$ crosses the $yz$-plane at $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$. Then the value of $(3a + 4b)$ is:

The equations of motion of a rocket are: $x = 2t, y = -4t, z = 4t$ where the time $t$ is given in seconds,and the coordinates of a moving point in kilometers. What is the path of the rocket? At what distance will the rocket be from the starting point $O(0, 0, 0)$ in $10$ seconds?

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