The distance of the point (1, 0, 2) from the | vedclass

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(A) Let the line be $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} = \lambda$.Then,any point on the line is given by $(3\lambda + 2, 4\lambda -

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$13$

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(A) Let the line be $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} = \lambda$.Then,any point on the line is given by $(3\lambda + 2, 4\lambda - 1, 12\lambda + 2)$.Since this point lies on the plane $x - y + z = 16$,we substitute these coordinates into the plane equation:$(3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2) = 16$$3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 16$$11\lambda + 5 = 16$$11\lambda = 11$$\lambda = 1$Substituting $\lambda = 1$ back into the coordinates,we get the point of intersection:$x = 3(1) + 2 = 5$$y = 4(1) - 1 = 3$$z = 12(1) + 2 = 14$So,the point of intersection is $(5, 3, 14)$.Now,we find the distance between the point $(1, 0, 2)$ and $(5, 3, 14)$ using the distance formula:$d = \sqrt{(5 - 1)^2 + (3 - 0)^2 + (14 - 2)^2}$$d = \sqrt{4^2 + 3^2 + 12^2}$$d = \sqrt{16 + 9 + 144}$$d = \sqrt{169} = 13$.Thus,the distance is $13$ units.

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

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