Let $P \text{and} Q$ 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 $R(1,2,3)$. If the centroid of the triangle $PQR$ is $\left(\alpha ,\beta ,\gamma \right),$ then ${\alpha }^2+{\beta }^2+{\gamma }^2$ is:

Let, $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}=\lambda$.

$\Rightarrow x=8\lambda -3, y=2\lambda +4, z=2\lambda -1$

Let, $P\equiv \left(8\lambda -3, 2\lambda +4, 2\lambda -1\right)$.

Now, given $R\left(1,2,3\right)$ and $PR=6 \& QR=6$.

So, by distance formula we get,

$\Rightarrow {\left(8\lambda -4\right)}^2+{\left(2\lambda +2\right)}^2+{\left(2\lambda -4\right)}^2=36$

$\Rightarrow 64{\lambda }^2+16-64\lambda +4{\lambda }^2+4+8\lambda +4{\lambda }^2+16-16\lambda =36$

$\Rightarrow 72{\lambda }^2-72\lambda =0$

$\Rightarrow \lambda =0, 1$

$\Rightarrow P\equiv \left(-3,4,-1\right) \text{and} Q\equiv \left(5,6,1\right)$

Hence, centroid of $△PQR$ will be,

$\left(\alpha ,\beta ,\gamma \right)\equiv \left(1,4,1\right)$.

$\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2=18$