When we calculate the shortest distance, call it $s$, of a point $Q$ with position vector $\bf{q}$ to a plane $\Pi$, we take the absolute value, so the orientation of our unit normal vector $\bar{\bf{n}}$ to the plane doesn't matter. But when we need to find the position vector $\bf{p}$ of a point $P$ on the plane that is closest to $Q$, how do we know that our normal is oriented the correct way? We can write for the position $\bf{p}$ $ = \bf{q}$ $ +s\bar{\bf{n}}$, but how do we know that will take us onto $\Pi$, and not to the "other side" of $Q$?
Edit. I suppose there's an easy way to check - simply plug in the resulting value of $\bf{p}$ into the equation of plane and see if it's satisfied; if not, then simply take the unit normal with an opposite sign. But is there a way to decide apriori?
It does not matter which way you chose the normal vector, the calculated position $\mathbf{p}$ will be the same.
If $\mathbf{p_0}$ is a point on the plane whose equation is
$\mathbf{n} \cdot (\mathbf{r} - \mathbf{p_0} ) = 0$
then
$ \mathbf{p} = \mathbf{q} - \mathbf{n} \dfrac{(\mathbf{q} - \mathbf{p_0}) \cdot \mathbf{n} }{\mathbf{n} \cdot \mathbf{n}}$
from this, you can see that if you replace $\mathbf{n}$ by $(\mathbf{-n})$ you get the same point $\mathbf{p}$.