Points, Lines, and Planes: Practice

All common points of the distinct planes form one line.

Since this line contains the distinct points $A$ and $B$, the unique-line foundation identifies it as $\overleftrightarrow{AB}$.

The common point $C$ must also lie on that line, so the three points are collinear.

Three noncollinear points determine exactly one plane. Both plane names must identify that same plane.

If the distinct coplanar lines had no common point, they would be parallel by definition. They therefore meet.

Distinct lines share at most one point, so the intersection is exactly one point.

The lines share $A$.

They are distinct, since otherwise $A$, $B$, and $C$ would be collinear.

Distinct lines share at most one point, so their intersection is only $A$.

Any plane containing lines $EF$ and $CG$ would contain the noncollinear points $E,F,G$, so it would be the top-face plane.

But $C$ is in the parallel bottom-face plane, outside the top-face plane. Thus no plane contains both lines, so they are skew.

The other pairs are respectively parallel, perpendicular, and parallel in the given rectangular prism.

Planes $\mathcal Q$ and $\mathcal R$ both contain every point of $\ell$.

Because they are distinct, all their common points form a line $k$. Choose two distinct points $U,V$ on $\ell$; they also lie on $k$.

The unique-line foundation makes $k$ and $\ell$ the same line, so the intersection is $\ell$.

Choose a point $X$ on $r$ other than $V$ and a point $Y$ on $s$ other than $V$.

Distinct lines have at most one common point, already $V$.

Thus $Y$ is off $r$, the unique line through $V$ and $X$, so $V,X,Y$ are noncollinear.

Their plane contains both lines by the line-in-plane foundation, so the lines are not skew. Coplanarity does not require perpendicularity.

Both rays in the correct statement start at $X$ and extend toward $Y$ and $Z$.

The other ray pairs point in opposite directions.

Segments $WX$ and $WZ$ have different endpoints, so they are different sets of points.

Choose distinct points $U$ and $V$ on $\ell$.

Point $T$ is off $\ell$, because $\ell$ lies in $\mathcal P$ while $T$ lies outside that plane.

The noncollinear points $U,V,T$ determine one plane. The line-in-plane foundation puts all of $\ell$ in that plane.

Any plane containing both $\ell$ and $T$ must contain the same noncollinear triple, so it must be this unique plane. It differs from $\mathcal P$.

Every point of $m$ belongs to the intersection, which is $k$.

Choose two distinct points on $m$. They also belong to $k$.

The unique-line foundation makes $m$ and $k$ the same line.

Three noncollinear points determine a unique plane. One point or one line can be shared by distinct planes, so neither of those guarantees coincidence.

Disjoint coplanar lines are parallel. Disjoint noncoplanar lines are skew. Without coplanarity information, both cases remain possible.