The distance, of the point $(7,-2,11)$ from the line

$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is :

If the shortest distance between the lines

$\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is :

Let the foot of perpendicular of the point $P(3,-2,-9)$ on the plane passing through the points $(-1,-2,-3),(9,3,4),(9,-2,1)$ be $Q(\alpha, \beta, \gamma)$. Then the distance of $Q$ from the origin is :

Let the system of linear equations

$-x+2 y-9 z=7$

$-x+3 y+7 z=9$

$-2 x+y+5 z=8$

$-3 x+y+13 z=\lambda$

has a unique solution $x=\alpha, y=\beta, z=\gamma$. Then the distance of the point

$(\alpha, \beta, \gamma)$ from the plane $2 x-2 y+z=\lambda$ is :