The point of intersection of lines $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1}$ and $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ is

Let $O$ be the origin,and $M$ and $N$ be the points on the lines $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$ and $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$ respectively such that $MN$ is the shortest distance between the given lines. Then $\overrightarrow{OM} \cdot \overrightarrow{ON}$ is equal to:

If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$,then the largest possible value of $|\lambda|$ is equal to ..........

The Cartesian equation of the line which is parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$ and passes through the point $(5, 2, -4)$ is . . . . . . .

If the angle between the lines,$\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{5 - x}{- 2} = \frac{7y - 14}{p} = \frac{z - 3}{4}$ is $\cos^{-1} \left( \frac{2}{3} \right)$,then $p$ is equal to

If the line passing through the points $(5, 1, a)$ and $(3, b, 1)$ crosses the $YZ$ plane at the point $(0, 17/2, -13/2)$,then $a+b=$