Distance between the parallel lines $\frac{x}{3}=\frac{y-1}{-2}=\frac{z}{1}$ and $\frac{x+4}{3}=\frac{y-3}{-2}=\frac{z+2}{1}$ is

The line joining points $(3, 5, -7)$ and $(-2, 1, 8)$ meets the $yz$-plane at which point?

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

The line $L$ given by $\frac{x - 2}{2} = \frac{y - 1}{b} = \frac{z + 1}{c}$ passes through the point $(1, 2, 3)$. Another line $K$ is parallel to line $L$ and has the equation $\frac{x + 2}{a} = \frac{y - 3}{2} = \frac{z + 4}{d}$. Then the distance between line $L$ and $K$ is

The equation of the line passing through the point $(0,1,2)$ and perpendicular to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is

If the shortest distance between the lines $\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$ and $\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$ is $\frac{38}{3 \sqrt{5}} k$ and $\int_0^{k}\left[x^2\right] dx=\alpha-\sqrt{\alpha}$,where $[x]$ denotes the greatest integer function,then $6 \alpha^3$ is equal to ............................