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(A) The direction vector of the line passing through $P(1, -2, 3)$ and $Q(5, -4, 7)$ is $\vec{v} = (5-1, -4-(-2), 7-3) = (4, -2, 4)$.The equation of t

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$155$

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(A) The direction vector of the line passing through $P(1, -2, 3)$ and $Q(5, -4, 7)$ is $\vec{v} = (5-1, -4-(-2), 7-3) = (4, -2, 4)$.The equation of the line is $\frac{x-1}{4} = \frac{y+2}{-2} = \frac{z-3}{4} = t$.Any point on this line is given by $(4t+1, -2t-2, 4t+3)$.The distance of this point from $P(1, -2, 3)$ is $\sqrt{(4t+1-1)^2 + (-2t-2+2)^2 + (4t+3-3)^2} = \sqrt{16t^2 + 4t^2 + 16t^2} = \sqrt{36t^2} = 6|t|$.Given the distance is $9$ units,$6|t| = 9$,so $t = \pm \frac{3}{2}$.For $t = \frac{3}{2}$,the point is $(4(\frac{3}{2})+1, -2(\frac{3}{2})-2, 4(\frac{3}{2})+3) = (7, -5, 9)$.For $t = -\frac{3}{2}$,the point is $(4(-\frac{3}{2})+1, -2(-\frac{3}{2})-2, 4(-\frac{3}{2})+3) = (-5, 1, -3)$.The distance of $(7, -5, 9)$ from the origin is $\sqrt{7^2 + (-5)^2 + 9^2} = \sqrt{49 + 25 + 81} = \sqrt{155}$.The distance of $(-5, 1, -3)$ from the origin is $\sqrt{(-5)^2 + 1^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35}$.Since the point is farther from the origin,we choose $(7, -5, 9)$.Thus,$\alpha^2 + \beta^2 + \gamma^2 = 7^2 + (-5)^2 + 9^2 = 49 + 25 + 81 = 155$.

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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