Given $P(0,1,0)\:and\:Q(0,0,1)$.
$\therefore PQ=\sqrt$
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Given equation of the plane is $x+y+z=3$
Normal to the plane is $\overrightarrow n=(1,1,1)$
Angle $\theta $ between the line $\overrightarrow {PQ}$ and the plane is given by
$sin\theta=\large\frac{\overrightarrow {PQ}.\overrightarrow n}{|\overrightarrow {PQ}||\overrightarrow b|}=\large\frac{(0,-1,1).(1,1,1)}{\sqrt 2. \sqrt 3}$$=0$
$\Rightarrow\:\theta=0$
Now
Projection of $PQ$ on the plane is $PQ\:cos\theta=PQ=\sqrt 2$