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 :

Any point on line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ <br/><br/>can be taken as $(8 \lambda-3,2 \lambda+4,2 \lambda-1)$ <br/><br/> If at a distance of 6 units from $R(1,2,3)$ <br/><br/>$$ \begin{aligned} & \Rightarrow(8 \lambda-3-1)^2+(2 \lambda+4-2)^2+(2 \lambda-1-3)^2=36 \\\\ & \left.\Rightarrow \lambda^2-\lambda=0 \text { \{on simplification }\right\} \\\\ & \Rightarrow \lambda=0, \lambda=1 \end{aligned} $$ <br/><br/>Here $P \& Q$ are $(-3,4,-1)$ and $(5,6,1)$ Centroid of $\triangle P Q R$ <br/><br/>$$ \begin{aligned} & (\alpha, \beta, \gamma) \equiv\left(\frac{5-3+1}{3}, \frac{6+4+2}{3}, \frac{1-1+3}{3}\right) \\\\ & \Rightarrow \alpha=1, \beta=4, \gamma=1 \\\\ & \Rightarrow \alpha^2+\beta^2+\gamma^2=18 \end{aligned} $$