Centroid and Distance Condition for Triangle PQR

To find the value of $\alpha + \beta + \gamma$, where $(\alpha, \beta, \gamma)$ is the centroid of $\Delta PQR$, we first determine the coordinates of the points $P$ and $Q$.

The given line on which $P$ and $Q$ lie is: $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2} = t$

Any general point $T$ on this line can be expressed in terms of the parameter $t$ as: $T = (8t - 3, 2t + 4, 2t - 1)$

Since $P$ and $Q$ lie on this line and are at a distance of $6$ units from the point $R(1, 2, 3)$, the square of the distance from $R$ to $T$ must be equal to $6^2 = 36$: $RT^2 = (8t - 3 - 1)^2 + (2t + 4 - 2)^2 + (2t - 1 - 3)^2 = 36$

Simplifying the expression: $(8t - 4)^2 + (2t + 2)^2 + (2t - 4)^2 = 36$

Expanding each term: $(64t^2 - 64t + 16) + (4t^2 + 8t + 4) + (4t^2 - 16t + 16) = 36$ $72t^2 - 72t + 36 = 36$ $72t^2 - 72t = 0$ $72t(t - 1) = 0$

This yields two values for $t$: $t = 0 \quad \text{and} \quad t = 1$

Substituting these values of $t$ back into the parametric equations gives the coordinates of $P$ and $Q$:

• For $t = 0$: $P = (-3, 4, -1)$
• For $t = 1$: $Q = (8(1) - 3, 2(1) + 4, 2(1) - 1) = (5, 6, 1)$

The coordinates of vertex $R$ are given as $(1, 2, 3)$.

The centroid $(\alpha, \beta, \gamma)$ of $\Delta PQR$ is given by the average of the coordinates of its vertices: $\alpha = \frac{x_P + x_Q + x_R}{3} = \frac{-3 + 5 + 1}{3} = 1$ $\beta = \frac{y_P + y_Q + y_R}{3} = \frac{4 + 6 + 2}{3} = 4$ $\gamma = \frac{z_P + z_Q + z_R}{3} = \frac{-1 + 1 + 3}{3} = 1$

Now, calculating $\alpha + \beta + \gamma$: $\alpha + \beta + \gamma = 1 + 4 + 1 = 6$

Thus, the correct option is C.