Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin and at distance of 9 units from the point $\mathrm{P}$, be $(\alpha, \beta, \gamma)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to :

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Aicharya · Accepted Answer

$P Q$ line $\begin{aligned} & \frac{x-1}{4}=\frac{y+2}{-2}=\frac{z-3}{4} \ & \operatorname{pt}(4 t+1,-2 t-2,4 t+3) \ & 	ext { distance }{ }^2=16 t^2+4 t^2+16 t^2=81 \ & t= \pm \frac{3}{2} \ & \operatorname{pt}(7,-5,9) \ & \alpha^2+\beta^2+\gamma^2=155 \end{aligned}$ option (1)

Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin…

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