If P(0, 1, 0) and Q(0, 0, 1) are two points, then the projection of $\vec{PQ}$ on the plane $x + y + z = 3 $ is

$\sqrt{2}$

The projection of PQ on the given plane is PQ cos θ, where θ is the angle between PQ and the plane.

Let $\vec{n}$ be a vector normal to the plane.

We have, $\vec{PQ} = -\hat{j} + \hat{k}$ and $\vec{n}=\hat{i} + \hat{i} + \hat{k}$

$∴ sin \theta = \frac{\vec{PQ}.\vec{n}}{|\vec{PQ}||\vec{n}|}= 0 ⇒ \vec{PQ}$ is parallel to the plane.

Hence, projection of $\vec{PQ}$ on teh given plane

$= |\vec{PQ}|cos \theta = |\vec{PQ}| = \sqrt{1+1}= \sqrt{2}$