Q:

Finding the closest point between a plane formed by a equation and another point

There's a question about the geometry on  the $ \mathbb{R}^{3} $ topic of my book that goes as follows:

Consider the plane $\pi$ : $ 2x + 3y - z = 20 $ and the point $P $(-4 ,2 ,6).

Find the point $Q$ under the $\pi$ plane that is closest to $P$ and find also the distance between $P$ and $Q$.
Now, first of all. How should I interpret a equation forming a plane? In that sense what does it mean to make the same equation equals to a number (20 in this example)? Like, if i can assign lots of values into $ x , y$ and $z$, wouldn't it form a bunch of planes instead of one plane? Second, how should i proceed in order to actually solve this type of exercises, at least the first steps or some intuition. Is the distance between the points $Q$ and $P$ related to the norm of a vector? Any tool i could use to better visualize how this would look in 3d? Thanks!

A:

The equation for the plane
$$\pi : 2x + 3y - z = 20$$
defines one and only one plane in $\mathbb R^3$. As you know, a plane is a 2-dimensional linear surface. You can think of this in the following way: You can pick any values for $x$ and $y$ as you like, but then $z$ has to be the specific value:
$$ z = 2x + 3y - 20$$
and in the same way, you could pick any value for $x$ and $z$, or $y$ and $z$, but then the last variable would have to be a specific value.
To the question itself. So, you want to find a point $Q = (q_1, q_2, q_3)$ on the plane $\pi$, i.e. you have $2q_1 + 3q_2 - q_3 = 20$ such that $Q$ is as close to $P$ as possible. As you mentioned, this is the same as minimizing the norm $\| Q - P\|$.
This can be done by projecting $P$ onto $\pi$.
You can visualize this by drawing it. Drawing a plane properly can be a bit tricky, but if you find a couple of points on the plane you can generally get a decent picture from there.

