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

If the two lines $l_{1}: \frac{x-2}{3}=\frac{y+1}{-2}, z=2$ and $l_{2}: \frac{x-1}{1}=\frac{2y+3}{\alpha}=\frac{z+5}{2}$ are perpendicular,then the angle between the lines $l_{2}$ and $l_{3}: \frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}$ is

The foot of the perpendicular from the point $A(2, 0, 5)$ to the line $\frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1}$ is $P(\alpha, \beta, \gamma)$. Then,which of the following is $NOT$ correct?

Let the point,on the line passing through the points $P(1, -2, 3)$ and $Q(5, -4, 7)$,farther from the origin and at a distance of $9$ units from the point $P$,be $(\alpha, \beta, \gamma)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:

The distance of the point $A(7, -2, 11)$ from the line $\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$ measured along the line $\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6}$ is:

When are the two lines $x = ay + b, z = cy + d$ and $x = a'y + b', z = c'y + d'$ perpendicular to each other?