Let a triangle PQR be such that P and Q lie on the line (x+3)/(8) =...

The equation of the given line is $\dfrac{x+3}{8} = \dfrac{y-4}{2} = \dfrac{z+1}{2}$.

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

Any point on this line can be taken as $S(4\lambda - 3, \lambda + 4, \lambda - 1)$.

Since the points $P$ and $Q$ lie on this line and are at a distance of $6$ units from $R(1, 2, 3)$, we have $RS^2 = 36$.

$(4\lambda - 3 - 1)^2 + (\lambda + 4 - 2)^2 + (\lambda - 1 - 3)^2 = 36$

$(4\lambda - 4)^2 + (\lambda + 2)^2 + (\lambda - 4)^2 = 36$

$16(\lambda^2 - 2\lambda + 1) + (\lambda^2 + 4\lambda + 4) + (\lambda^2 - 8\lambda + 16) = 36$

$18\lambda^2 - 36\lambda + 36 = 36$

$18\lambda(\lambda - 2) = 0 \Rightarrow \lambda = 0 \text{ or } \lambda = 2$

For $\lambda = 0$, the point is $P(-3, 4, -1)$.

For $\lambda = 2$, the point is $Q(5, 6, 1)$.

The centroid $(\alpha, \beta, \gamma)$ of $\triangle PQR$ is given by:

$\alpha = \dfrac{-3 + 5 + 1}{3} = 1$

$\beta = \dfrac{4 + 6 + 2}{3} = 4$

$\gamma = \dfrac{-1 + 1 + 3}{3} = 1$

Therefore, $\alpha + \beta + \gamma = 1 + 4 + 1 = 6$.

Answer: $6$