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

The given line can be written in parametric form by putting the common ratio equal to $$t$$:

$$x = 8t - 3,\;\; y = 2t + 4,\;\; z = 2t - 1$$

Let a general point on this line be $$P(t)\bigl(8t - 3,\;2t + 4,\;2t - 1\bigr)$$.

The fixed vertex of the triangle is $$R(1,\,2,\,3)$$. Distance formula tells us

$$PR^2 = (8t - 3 - 1)^2 + (2t + 4 - 2)^2 + (2t - 1 - 3)^2$$ $$\;\;\;= (8t - 4)^2 + (2t + 2)^2 + (2t - 4)^2$$

Expanding each term:

$$\begin{aligned} (8t - 4)^2 &= 64t^2 - 64t + 16,\\ (2t + 2)^2 &= 4t^2 + 8t + 4,\\ (2t - 4)^2 &= 4t^2 - 16t + 16. \end{aligned}$$

Adding them gives

$$PR^2 = 72t^2 - 72t + 36.$$

The problem states $$PR = 6$$, hence $$PR^2 = 36$$:

$$72t^2 - 72t + 36 = 36 \;\;\Longrightarrow\;\; 72t^2 - 72t = 0$$ $$\Longrightarrow\; 72t(t - 1) = 0$$

Therefore $$t = 0$$ or $$t = 1$$. So the two required points on the line are

For $$t = 0$$: $$P(-3,\,4,\,-1)$$
For $$t = 1$$: $$Q(5,\,6,\,1)$$

Coordinates of all three vertices are now known:

$$P(-3,4,-1),\; Q(5,6,1),\; R(1,2,3).$$

The centroid $$G(\alpha,\beta,\gamma)$$ of $$\triangle PQR$$ is given by the component-wise average:

$$\alpha = \frac{-3 + 5 + 1}{3},\; \beta = \frac{4 + 6 + 2}{3},\; \gamma = \frac{-1 + 1 + 3}{3}.$$

Hence

$$\alpha + \beta + \gamma = \frac{(-3 + 5 + 1) + (4 + 6 + 2) + (-1 + 1 + 3)}{3} = \frac{18}{3} = 6.$$

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

Option C which is: $$6$$