The distance of the point (1, 1, 9) from the | vedclass

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(D) Let the line be $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2} = k$. Then,$x = k+3$,$y = 2k+4$,and $z = 2k+5$. Substitute these into the plane equ

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$\sqrt{38}$

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(D) Let the line be $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2} = k$. Then,$x = k+3$,$y = 2k+4$,and $z = 2k+5$. Substitute these into the plane equation $x+y+z=17$: $(k+3) + (2k+4) + (2k+5) = 17$. $5k + 12 = 17 \implies 5k = 5 \implies k = 1$. Substituting $k=1$ into the coordinates,we get the point of intersection $P = (1+3, 2(1)+4, 2(1)+5) = (4, 6, 7)$. The distance between $(1, 1, 9)$ and $(4, 6, 7)$ is given by the distance formula: $d = \sqrt{(4-1)^2 + (6-1)^2 + (7-9)^2} = \sqrt{3^2 + 5^2 + (-2)^2} = \sqrt{9 + 25 + 4} = \sqrt{38}$.

The distance of the point $(1, 1, 9)$ from the point of intersection of the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ and the plane $x+y+z=17$ is

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