Geometrically Construct a Plane to Create Cross Section or Finding the Cross Section with Linear Algebra

Question

I wanted to ask this question so badly because i dont find any answer on google for it.

1. How do we construct cross section that's formed inside a 3d figures (especially cube) (or maybe i should say how do we geometrically construct a plane inside 3d figure that's defined by 3 points)

2. What are the geometrical reasoning, and what is allowed to do and not allowed in order to construct a cross section ?

3.Lastly can we somehow know the shape of cross section with linear algebra (like finding the intersection point, but without plugging in x/y/z value one by one in plane equation) ? .

Actually i have seen 2 question in this site that also ask this, but they didn't explain the geometrical reasoning of why those action are legal to do so i'm a bit confused. Please dont introduce some high level college math because im not even in college yet.

Take this example : construct a plane that goes through point P,Q,R and determine its cross section with the cube !

Cube with side 6 unit

Answer 1

The general idea is that we can construct an intersection of lines if they lie on the same plane.

1. We start by constructing projections $R'$ and $P'$ of $R$ and $P$ on a plane $ABCD$. We can do it by drawing line parallel to $AE$ (which is perpendicular to $ABCD$ because it's cube) through points $R$ and $P$ and then finding intersections with lines $AD$ and $BC$. The reasoning is the following. Plane $ADHE||ABCD$ and $R\in ADHE$, so $RR'$ should lie on $ADHE$. Since $R'\in ABCD$ and $R'\in ADHE$, then it lies on intersection of those planes, line $AD$. The same applies to $PP'$.

2. Since $RR'$ and $PP'$ are parallel, we can construct a plane $RR'PP'$. It should obviously contain lines $RP$ and $R'P'$. The intersection of those lines, point $S$, lies on plane $ABCD$ and also lies on section plane $PQR$.

3. Line $QS$ lies both in plane $PQR$ and in plane $ABCD$, so it is the intersection of those planes. Intersection of $QS$ with lines $BC$ and $AD$ give points $T$ and $U$.

4. Since $PT$ lies on $BCGF$, it can be intersected with $BF$ and $GC$. The same true for $UR$ and $HD$. We got all the needed intersections of $PQR$ with edges of cube.

5. Finally, we connect dots.

Finding the numerical values of intersection is just a thorough study of proportions. For example, $SP':SR'=PP':RR'=6:3=1:2$. Since $DR'=CP'=3$, then $DC$ and $R'P'$ intersect in middle points and $\angle P'R'D=45^\circ$. And so on.