Q.Show that lines x = y, z = 0 and x + y = 0, z = 0 intersect each other. Find the vector equation of the plane determined by them.

[!TLDR] Both lines pass through the origin in the $z=0$ plane, so they intersect there; the plane containing them is $z=0$. …

Line 1: $x=y,\ z=0$, which can be parametrised as $(t,t,0)$ — it passes through the origin $(0,0,0)$ (at $t=0$) with direction $(1,1,0)$.

Line 2: $x+y=0,\ z=0$, parametrised as $(s,-s,0)$ — it also passes through the origin (at $s=0$) with direction $(1,-1,0)$.

Both lines pass through the common point $(0,0,0)$, so they intersect there (and since their directions $(1,1,0)$ and $(1,-1,0)$ are not parallel, this is a genuine single point of intersection).

Both direction vectors have zero $z$-component and both lines pass through a point with $z=0$, so both lines lie entirely within the plane $z=0$. Equivalently, the normal is …

Parametrise each line to find a point and direction; a shared point already lying on both lines proves intersection directly. For the plane, either reason geometrically (both lines lie in $z=0$ since both have …

Overlooking that both lines pass through the very same point, the origin (it is easy to only check that the lines are coplanar via the determinant test and forget the simpler direct observation that they share a point); forgetting to note the direction vectors …