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(C) Since the direction cosines of the line $PQ$ are equal and positive,let them be $l, m, n$. Since $l=m=n$ and $l^2+m^2+n^2=1$,we have $3l^2=1$,so $

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$\sqrt{3}$ units

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(C) Since the direction cosines of the line $PQ$ are equal and positive,let them be $l, m, n$. Since $l=m=n$ and $l^2+m^2+n^2=1$,we have $3l^2=1$,so $l=m=n=\frac{1}{\sqrt{3}}$.The direction ratios of the line $PQ$ are proportional to $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$,which can be taken as $1, 1, 1$.The equation of the line $PQ$ passing through $P(2,-1,2)$ is $\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = k$.Thus,any point on the line is $(k+2, k-1, k+2)$.Since this point $Q$ lies on the plane $2x+y+z=9$,we have $2(k+2) + (k-1) + (k+2) = 9$.$4k + 5 = 9$ $\Rightarrow 4k = 4$ $\Rightarrow k = 1$.Substituting $k=1$,the coordinates of $Q$ are $(3, 0, 3)$.The length $PQ = \sqrt{(3-2)^2 + (0-(-1))^2 + (3-2)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ units.

$A$ line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. Then the length of the line segment $PQ$ equals

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