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(D) The equation of the line passing through $(1, -5, 9)$ and parallel to the line $x = y = z$ is given by $\frac{x - 1}{1} = \frac{y + 5}{1} = \frac{

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$10\sqrt{3}$

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(D) The equation of the line passing through $(1, -5, 9)$ and parallel to the line $x = y = z$ is given by $\frac{x - 1}{1} = \frac{y + 5}{1} = \frac{z - 9}{1} = \lambda$.Any point on this line is of the form $P(\lambda + 1, \lambda - 5, \lambda + 9)$.Since this point $P$ lies on the plane $x - y + z = 5$,we substitute the coordinates into the plane equation:$(\lambda + 1) - (\lambda - 5) + (\lambda + 9) = 5$$\lambda + 1 - \lambda + 5 + \lambda + 9 = 5$$\lambda + 15 = 5$$\lambda = -10$.Substituting $\lambda = -10$ back into the coordinates of $P$,we get:$P = (-10 + 1, -10 - 5, -10 + 9) = (-9, -15, -1)$.The distance between the points $(1, -5, 9)$ and $(-9, -15, -1)$ is calculated using the distance formula:$d = \sqrt{(-9 - 1)^2 + (-15 - (-5))^2 + (-1 - 9)^2}$$d = \sqrt{(-10)^2 + (-10)^2 + (-10)^2}$$d = \sqrt{100 + 100 + 100} = \sqrt{300} = 10\sqrt{3}$.

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is

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