<s>[INST] Finding K such that a line is perpendicular or parallel to a given plane? Known information:
Plane $P = 3x -5y + 2z =1 (R^3)$
Parametrics:
 x = 5 + t
 y = 2 - 2t
 z = -1 - 6t

I've solved for $t$, $t = -2$. Point of intersection: $(3,6,11)$
The part causing problems is, given vectors:
 A = (2, -7, -6)
 B = (-4, 3, k)

Vector of $A$ and $B = AB: (-8, -21, -6k)$
Solve for $k$ such that vector $AB$ is perpendicular to the plane, and for $k$ such that $AB$ is parallel to the plane.
Thank you very much for any help. I've been pouring over notes for how to do this, with little success. Thank you.
Also, please note: the accuracy of vector $AB$ is not 100%. IF the math is wrong, that may be part of the issue, so please note that.
 [/INST] Your plane is described by
$$
(3, -5, 2)  \cdot (x,y,z) = 1 \iff \\
\frac{1}{\lVert (3,-5,2) \rVert} (3, -5, 2)  \cdot (x,y,z) =
\frac{1}{\lVert (3,-5,2) \rVert} \\
\frac{1}{\sqrt{38}} (3, -5, 2)  \cdot (x,y,z) =
\frac{1}{\sqrt{38}} \iff \\
n \cdot u = d
$$
where $n$ is a unit normal vector and $d$ is the signed distance to the origin.
a) Assuming $AB = B - A = (-6, 10, k + 6)$ then a vector perpendicular to the plane is parallel to $n$:
$$
\alpha n = AB \iff \\
\alpha \frac{1}{\sqrt{38}} (3, -5, 2) = (-6, 10, k + 6)
$$
where $\alpha \in \mathbb{R}$ is some scalar. Using $\alpha = -2 \sqrt{38}$ we get 
$$
(-6, 10, -4) = (-6, 10, k + 6) \Rightarrow \\\
-4 = k + 6 \iff \\
k = -10
$$
b) If $AB$ is parallel to the plane it is perpendicular to $n$:
$$
0 
= n \cdot (-6, 10, k+6) \\
= \frac{1}{\sqrt{38}} (3, -5, 2) \cdot (-6, 10, k+6) \\
= \frac{1}{\sqrt{38}} (-18 -50 + 2k + 12) \\
= \frac{1}{\sqrt{38}} (2k - 56) \iff \\
0 = 2k - 56 \iff \\
2k = 56 \iff \\
k = 28
$$
</s>