Find the coordinates of the foot of the perpendicular from the point $(2, 4, 1)$ to the line $\vec{r} = (-5, -3, 6) + k(1, 4, -9)$,where $k \in R$.

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}$ is

The square of the distance of the point $(-2, -8, 6)$ from the line $\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}$ along the line $\frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$ is equal to:

Consider the line $L$ passing through the points $(1, 2, 3)$ and $(2, 3, 5)$. The distance of the point $A\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ from the line $L$ along the line $\frac{3x-11}{2} = \frac{3y-11}{1} = \frac{3z-19}{2}$ is equal to:

The shortest distance between the lines $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ lies in the interval

Consider the lines $L_1: x-1=y-2=z$ and $L_2: x-2=y=z-1$. Let the feet of the perpendiculars from the point $P(5,1,-3)$ on the lines $L_1$ and $L_2$ be $Q$ and $R$ respectively. If the area of the triangle $PQR$ is $A$,then $4A^2$ is equal to: