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 :

If $(\alpha, \beta)$ is the orthocenter of the triangle $\mathrm{ABC}$ with vertices $A(3,-7), B(-1,2)$ and $C(4,5)$, then $9 \alpha-6 \beta+60$ is equal to :

Let $A_{1}$ and $A_{2}$ be two arithmetic means and $G_{1}, G_{2}, G_{3}$ be three geometric

means of two distinct positive numbers. Then $G_{1}^{4}+G_{2}^{4}+G_{3}^{4}+G_{1}^{2} G_{3}^{2}$ is equal to :

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 :