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 a distance of 9 units from the point P, be (\alpha, \beta, \gamma). Then \alpha^2+\beta^2+\gamma^2 is equal to :

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Answer: B. 155 Line through P Q \frac{x-1}{4}=\frac{y+2}{-2}=\frac{z-3}{4} Any point on P Q. be R(4 \lambda+1,-2 \lambda-2,4 \lambda+3) P R=9 unit (P R)^2=81 (4 \lambda+1-1)^2+(-2 \lambda-2+2)^2+(4 \lambda+3-3)^2=81 16 \lambda^2+4 \lambda^2+16 \lambda^2=81 36 \lambda^2=81 \lambda= \pm \frac{9}{6}= \pm \frac{3}{2} 	herefore R can be (7,-5,9) or (-5,1,-3) Distance from origin for both points be \sqrt{49+25+81} and \sqrt{25+1+9}=\sqrt{35} 	herefore Distance of (7,-5,9) is farthest from origin 	herefore(\alpha, \beta, \gamma)=(7,-5,9) Now 7^2+(-5)^2+9^2=155

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 a distance of 9 units from the point $$P$$, be $$(\alpha, \beta, \gamma)$$. Then $$\alpha^2+\beta^2+\gamma^2$$ is equal to :

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