Let PandQ be the points on the line (x + 3)/(8) = (y - 4)/(2) = (z...

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

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

Let, $P ≡ \left(8\lambda - 3, 2\lambda + 4, 2\lambda - 1\right).</p><p>Now, given$R\left(1, 2, 3\right)$and PR = 6 \text{ and } QR = 6$.

So, by distance formula we get,

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

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

$⇒ 72{\lambda}^{2} - 72\lambda = 0$

$⇒ \lambda = 0, 1$

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

Hence, centroid of $△ PQR$ will be,

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

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