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

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(D) Let the line be $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2} = t$.Then,$x = t+3$,$y = 2t+4$,and $z = 2t+5$.For the point of intersection with th

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$\sqrt{38}$

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(D) Let the line be $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2} = t$.Then,$x = t+3$,$y = 2t+4$,and $z = 2t+5$.For the point of intersection with the plane $x+y+z=17$,substitute these coordinates into the plane equation:$(t+3) + (2t+4) + (2t+5) = 17$$5t + 12 = 17$$5t = 5 \Rightarrow t = 1$.Substituting $t=1$ back into the line equations,the point of intersection is $(1+3, 2(1)+4, 2(1)+5) = (4, 6, 7)$.The distance between the points $(1, 1, 9)$ and $(4, 6, 7)$ is given by the distance formula:$d = \sqrt{(4-1)^2 + (6-1)^2 + (7-9)^2}$$d = \sqrt{3^2 + 5^2 + (-2)^2}$$d = \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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