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 vector equation of the line $2x+4=3y+1=6z-3$ is

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:

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 symmetric form of the equation of the line $x = ay + b$ and $z = cy + d$ is

The angle between the straight lines $\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z + 3}{4}$ and $\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{-3}$ is ......... $^o$