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(B) The equation of the line passing through $(1, -2, 3)$ and parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{-6}$ is given by $\frac{x-1}{

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$1$

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(B) The equation of the line passing through $(1, -2, 3)$ and parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{-6}$ is given by $\frac{x-1}{2} = \frac{y+2}{3} = \frac{z-3}{-6} = r$.Any point on this line is $(2r+1, 3r-2, -6r+3)$.Since this point lies on the plane $x - y + z = 5$,we substitute these coordinates into the plane equation:$(2r+1) - (3r-2) + (-6r+3) = 5$.$2r + 1 - 3r + 2 - 6r + 3 = 5$.$-7r + 6 = 5$.$-7r = -1$.$r = \frac{1}{7}$.The distance between the point $(1, -2, 3)$ and the intersection point $(2r+1, 3r-2, -6r+3)$ is $\sqrt{(2r)^2 + (3r)^2 + (-6r)^2} = \sqrt{4r^2 + 9r^2 + 36r^2} = \sqrt{49r^2} = 7|r|$.Substituting $r = \frac{1}{7}$,the distance is $7 	imes \frac{1}{7} = 1$.

The distance of the point $(1, -2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{-6}$ is

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