Is there a relationship between the existence of parallel vectors on two planes, and their line of intersection. Let me state the context first:
I have a question which goes like this:
We have two planes:
$$\pi_1 : r = (2,1,1)^{\top} + \lambda(-2,1,8)^{\top} + \mu(1, -3, -9)^{\top}$$
$$\pi_2 : r = (2,0,1)^{\top} + s(1,2,1)^{\top} + t(1,1,1)^{\top}$$
I have to show that for all $p \in \pi_2 \cup \pi_2$, when we express $p$ as an element of $\pi_1$ using it's equation the we have $\lambda = \mu$. 
Notice that the equation of $\pi_1$ gives us two vectors which lie along $\pi_1$, i.e. $(-2,1,8)^{\top}$ and $(1, -3, -9)^{\top}$, when we add these we obtain $(-1,-2,-1)^{\top}$, which is parallel to $\pi_2$, since it's one of the vectors specified in the equation for $\pi_2$, just flipped over. Also, notice that $(2,1,1)^{\top}$ lies on $\pi_2$ with $s = 1$ and $t = -1$. Intuitively these two conditions imply that the line of intersection lies along the sum of the basis vectors for $\pi_1$, so any point on this line satisfies $\lambda = \mu$. Here's a conjecture that attempts to formalise this intuition:
If we have $v \in \mathbb{R}^3$, $\pi_1$ and $\pi_2$ as planes in $\mathbb{R}^3$, and $p \in \pi_1 \cap \pi_2$ such that $p + v \in \pi_1 \cap \pi_2$, then the line of intersection of $\pi_1$ and $\pi_2$ has the direction vector $v$. I think I can draw a picture which heuristically justifies this, how I justify it algebraically?
 A: Yes, your conjecture is true. The crucial fact of note here is that a plane is an affine subspace - that is to say, it satisfies the following definition:

A subset $S$ of a vector space is an affine subspace if, for any pair of vectors $s_1,s_2\in S$ and any pair of numbers $a$ and $b$ with $a+b=1$, it holds that $as_1+bs_2\in S$

Intuitively, this says, "If an affine subspace contains two points, it contains the line between those two points too" since the condition that $a+b=1$ defines a sort of weighted average between (or beyond) the points. You can prove that if $s_1$ is a point on the plane parameterized as $(\lambda_1,\mu_1)$ and $s_2$ is a point on the plane parametrized as $(\lambda_2,\mu_2)$ then, so long as $a+b=1$, we have the $as_1+bs_2$ will be parameterized as $(a\lambda_1+b\lambda_2,a\mu_1+b\mu_2)$, which suffices to show that a plane is an affine subspace. You could also note that a plane can be written as the set of vectors $v$ such that $f(v)=c$ for some linear function $f$ and a constant $c$, and then given points $s_1$ and $s_2$ on the plane, we can write $$f(as_1+bs_2)=af(s_1)+bf(s_2)=ac+bc=(a+b)c=c$$
to show that affine combinations are on the plane too.
From here, it's easy: Clearly, the intersection of two affine subspaces is an affine subspace (as tends to be the case when we have condition of "A subset closed under ____ operation"). So, if $p$ and $p+v$ are the affine subspace which is the intersection of the two planes, so is any affine sum thereof - which is exactly the points of the form $p+\alpha v$. Assuming the planes are distinct (i.e. don't intersect everywhere), then their intersection would be parameterized by $p+\alpha v$.
