The perpendicular distance of point (1, 1, 3) on the line $\frac{x}{2}=\frac{y+1}{3}=\frac{-z-3}{-5}$ is

The image of the point (1, 6, 3) with respect to the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ is

The direction ratios of the line perpendicular to the lines $\dfrac{x-5}{2} = \dfrac{y+11}{-3} = \dfrac{z+3}{1}$and $\dfrac{x-7}{1} = \dfrac{y+2}{2} = \dfrac{z-4}{-2}$ are proportional to:

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:

Distance of the point $(2, 4, -1)$ from the line $\frac{10+2x}{2} = \frac{y+3}{4} = \frac{6-z}{9}$ is

Match List-I with List-II

Choose the correct answer from the options given below:

The angle at which the line, $\frac{x-1}{0} = \frac{2-y}{-1} = \frac{2z-3}{-2}$ is inclined with the positive direction of z-axis is