Question
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 $\Delta PQR$, then $\alpha + \beta + \gamma$ is equal to:
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Solution
Sol. [Key Concept: Parametrise P on the line; $PR = 6$ fixes the parameter, then the centroid is the average of P, Q, R.]
Let the point P is $(8\lambda - 3,2\lambda + 4,2\lambda - 1)$
$${PR = 6 \Rightarrow {PR}^{2} = 36 }{(8\lambda - 4)^{2} + (2\lambda + 2)^{2} + (2\lambda - 4)^{2} = 36 }{(4\lambda - 2)^{2} + (\lambda + 1)^{2} + (\lambda - 2)^{2} = 9 }{\lambda^{2} - \lambda = 0 \Rightarrow \lambda = 0,1 }{\lambda = 0 \Rightarrow P( - 3,4, - 1) }{\lambda = 1 \Rightarrow P(5,6,1) }{R(1,2,3) }$$G(1,4,1)
$$= \alpha + \beta + \gamma = 6 $$
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