The value of $\lambda$ so that the lines $\dfrac{1-x}{3} = \dfrac{7y-14}{2\lambda} = \dfrac{z-3}{2}$ and $\dfrac{7-7x}{3\lambda} = \dfrac{y-5}{1} = \dfrac{6-z}{5}$

are at right angles is:

If the lines $\frac{1-x}{2}=\frac{2y}{p}=\frac{z-1}{1}$ and $\frac{3-2x}{p}=\frac{y-1}{2}=\frac{z}{1}$ are parallel then value of 'p' is

The shortest distance between lines $L_1:\frac{x+3}{-4}=\frac{6-y}{-3}=\frac{z}{2}$ and $L_2:\frac{x+2}{-4}=\frac{y}{1}=\frac{z-7}{1}$ is:

If the lines $\frac{1-x}{3} = \frac{y-2}{2\lambda} = \frac{z-3}{2}$ and $\frac{x-1}{3\lambda} = \frac{y-1}{1} = \frac{6-z}{5}$ are perpendicular, then $\lambda$ is equal to

The angle between the lines

$l_1: \frac{x + 1}{1} = \frac{2 - y}{2} = \frac{z - 1}{1}$ and $l_2: \frac{x - 1}{4} = \frac{2y - 4}{6} = \frac{z - 1}{2}$ is

The vector equation of line passing through $(-1, 3, -2)$ and perpendicular to the lines $\frac{x+4}{1} = \frac{y}{2} = \frac{z-3}{3}$ and $\frac{x+2}{-3} = \frac{y+5}{2} = \frac{z-6}{5}$ is

If the line $\frac{-x+1}{3} = \frac{-y-2}{-2k} = \frac{z+3}{2}$ and $\frac{-1+x}{3k} = \frac{-1 +y}{1} = \frac{-z+6}{5}$ are perpendicular, then the value of k is: