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

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(A) 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} = \lam

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$2 \sqrt{14}$

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(A) The equation of the line passing through $P(2, -3, 4)$ and $Q(8, 0, 10)$ is given by: $\frac{x-2}{8-2} = \frac{y+3}{0+3} = \frac{z-4}{10-4} = \lambda$ $\frac{x-2}{6} = \frac{y+3}{3} = \frac{z-4}{6} = \lambda$ This gives the coordinates of any point on the line as $(6\lambda + 2, 3\lambda - 3, 6\lambda + 4)$. Since point $R(4, y, z)$ lies on this line,we equate the $x$-coordinate: $6\lambda + 2 = 4 \Rightarrow 6\lambda = 2 \Rightarrow \lambda = \frac{1}{3}$. Now,find $y$ and $z$: $y = 3(\frac{1}{3}) - 3 = 1 - 3 = -2$ $z = 6(\frac{1}{3}) + 4 = 2 + 4 = 6$ So,the coordinates of $R$ are $(4, -2, 6)$. The distance of $R(4, -2, 6)$ from the origin $(0, 0, 0)$ is: $d = \sqrt{4^2 + (-2)^2 + 6^2} = \sqrt{16 + 4 + 36} = \sqrt{56} = \sqrt{4 	imes 14} = 2\sqrt{14}$.

If a point $R(4, y, z)$ lies on the line 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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