The $xy$-plane divides the line segment joining the points $(2, 4, 5)$ and $(-4, 3, -2)$ in the ratio:

Find the foot of the perpendicular from the point $(2, 3, -8)$ to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$. Also,find the perpendicular distance from the given point to the line.

The coordinates of the foot of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{6 - x}{-3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ are:

The lines $x = ay + b, z = cy + d$ and $x = a'y + b', z = c'y + d'$ are perpendicular to each other,if

The Cartesian equation of a line is $\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$. Find the vector equation for the line.

Let $Q(a,b,c)$ be the image of the point $P(3,2,1)$ in the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}.$ Then the distance of $Q$ from the line $\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$ is