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 :

Let $\mathrm{S}$ be the set of all values of $\lambda$, for which the shortest distance between

the lines $\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$ is 13. Then $8\left|\sum\limits_{\lambda \in S} \lambda\right|$ is equal to :