Let $\mathrm{P}$ and $\mathrm{Q}$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of 6 units from the point $\mathrm{R}(1,2,3)$. If the centroid of the triangle PQR is $(\alpha, \beta, \gamma)$, then $\alpha^2+\beta^2+\gamma^2$ is :

If the mirror image of the point $P(3,4,9)$ in the line

$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$, then 14 $(\alpha+\beta+\gamma)$ is :

If the shortest distance between the lines

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