JEE MAIN - Mathematics (2024 - 4th April Morning Shift - No. 10)
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 :
150
155
160
165
Explanation
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$$(P R)^2=81$$
$$(4 \lambda+1-1)^2+(-2 \lambda-2+2)^2+(4 \lambda+3-3)^2=81$$
$$16 \lambda^2+4 \lambda^2+16 \lambda^2=81$$
$$36 \lambda^2=81$$
$$\lambda= \pm \frac{9}{6}= \pm \frac{3}{2}$$
$$\therefore R$$ can be $$(7,-5,9)$$ or $$(-5,1,-3)$$
Distance from origin for both points be $$\sqrt{49+25+81}$$ and $$\sqrt{25+1+9}=\sqrt{35}$$
$$\therefore$$ Distance of $$(7,-5,9)$$ is farthest from origin
$$\therefore(\alpha, \beta, \gamma)=(7,-5,9)$$
Now $$7^2+(-5)^2+9^2=155$$
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