If a point R(4, y, z) lies on the line segmen | vedclass

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(C) The equation of the line passing through $P(2, -3, 4)$ and $Q(8, 0, 10)$ is given by $\frac{x-2}{8-2} = \frac{y-(-3)}{0-(-3)} = \frac{z-4}{10-4}$.

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$2\sqrt{14}$

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(C) The equation of the line passing through $P(2, -3, 4)$ and $Q(8, 0, 10)$ is given by $\frac{x-2}{8-2} = \frac{y-(-3)}{0-(-3)} = \frac{z-4}{10-4}$.This simplifies to $\frac{x-2}{6} = \frac{y+3}{3} = \frac{z-4}{6}$.Since point $R(4, y, z)$ lies on this line,we substitute $x=4$ into the equation:$\frac{4-2}{6} = \frac{y+3}{3} = \frac{z-4}{6}$.$\frac{2}{6} = \frac{1}{3} = \frac{y+3}{3} = \frac{z-4}{6}$.From $\frac{1}{3} = \frac{y+3}{3}$,we get $y+3 = 1$,so $y = -2$.From $\frac{1}{3} = \frac{z-4}{6}$,we get $z-4 = 2$,so $z = 6$.Thus,the coordinates of $R$ are $(4, -2, 6)$.The distance of $R(4, -2, 6)$ from the origin $(0, 0, 0)$ is $\sqrt{4^2 + (-2)^2 + 6^2} = \sqrt{16 + 4 + 36} = \sqrt{56} = 2\sqrt{14}$.

If a point $R(4, y, z)$ lies on the line segment joining the points $P(2, -3, 4)$ and $Q(8, 0, 10)$,then the distance of $R$ from the origin is

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