Line parallel to plane but not on the plane? A question I'm trying to solve, given line:
$$x= t+1,y=t+b,z=ct+3$$
Determine all the pairs of the values of parameters $b$ and $c$ so that the line is parallel to the following plane but it isn't on the plane:
$$x+y+z=3$$
How should I get started with this? Also, can anyone link me to a good resource to learn more about this topic (parallel/perpendicular lines/planes involving equations) because I keep seeing them in my exercises but my professor never really bothered explaining them.
 A: Subs $$x= t+1, \quad y=t+b, \quad z=ct+3$$ into the equation of the plane, we have
$$(2+c)t + b+1 = 0.$$
The line parallel to the plane when and only when the equation has no solution, and then $2 +c = 0$ and $b+1 \ne 0$. That is mean $c = -2$ and $b \ne  -1$. 
A: There are several ways to answer this question. The first and perhaps easier way is to check if the line is orthogonal to the normal vector of the plane. The normal vector can be read off of the scalar equation as 
$$\hat{n} = \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}$$ 
while your line can be written as 
$$\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}1 \\ 1 \\ c\end{pmatrix}t + \begin{pmatrix}1 \\ b \\ 3\end{pmatrix}$$
First I want you to notice two things. First, it is $c$ which determines the direction of the line, $c$ is the only parameter which determines if the line will be parallel to the plane. On the other hand, $b$ is the factor which determines the spatial location of the line, changing $b$ will move the line around but it will not alter it's direction. 
So first, let us determine the direction of the line. We wish for the direction vector of the line to be orthogonal to the normal vector of the plane, to do that, we must have
$$\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}\cdot \begin{pmatrix}1 \\ 1 \\ c\end{pmatrix} = 1 + 1 + c = 0$$
It is clear that we require $c=-2$. Now that we know the direction for which the line is parallel, we can determine some $b$ for which the line isn't on the plane. Unlike $c$, the choice of $b$ will not be unique. In fact, it's enough to choose any $b$ such that
$$\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}1 \\ b \\ 3\end{pmatrix}$$
does not satisfy the equation of the plane. Therefore we need
$$x + y + z = 1 + b + 3 \neq 3$$
and we can see that we can take any $b$ except for $b=-1$.
