Let P and Q be the points on the line frac{x+ | vedclass

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(C) The general point on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2} = \lambda$ is given by $P, Q = (8\lambda-3, 2\lambda+4, 2\lambda-1)$.The

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$18$

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(C) The general point on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2} = \lambda$ is given by $P, Q = (8\lambda-3, 2\lambda+4, 2\lambda-1)$.The distance from this point to $R(1,2,3)$ is $6$ units,so the square of the distance is $36$:$(8\lambda-3-1)^2 + (2\lambda+4-2)^2 + (2\lambda-1-3)^2 = 36$$(8\lambda-4)^2 + (2\lambda+2)^2 + (2\lambda-4)^2 = 36$$64(\lambda^2 - \lambda + \frac{1}{4}) + 4(\lambda^2 + 2\lambda + 1) + 4(\lambda^2 - 4\lambda + 4) = 36$$64\lambda^2 - 64\lambda + 16 + 4\lambda^2 + 8\lambda + 4 + 4\lambda^2 - 16\lambda + 16 = 36$$72\lambda^2 - 72\lambda + 36 = 36$$72\lambda(\lambda - 1) = 0$Thus,$\lambda = 0$ or $\lambda = 1$.For $\lambda = 0$,the point is $P(-3, 4, -1)$.For $\lambda = 1$,the point is $Q(5, 6, 1)$.The centroid of $\Delta PQR$ is $(\frac{-3+5+1}{3}, \frac{4+6+2}{3}, \frac{-1+1+3}{3}) = (1, 4, 1) = (\alpha, \beta, \gamma)$.Therefore,$\alpha^2 + \beta^2 + \gamma^2 = 1^2 + 4^2 + 1^2 = 1 + 16 + 1 = 18$.

Let $P$ 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 $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2$ is:

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