Describing the plane given by a line and a vector I'm studying lines and planes and am stumped by the following problem:

If $L$ is the line through the point $A = (3,2,1)$ and parallel to the
vector $v = (-2,1,3)$, what's the equation of the plane that contains
$L$ and the point $B = (-2,3,1)$?

I'm trying to describe the plane in terms of $A$, $v$, and $B$ and then find a normal vector. The line $L$ is the span of $v$ shifted by $A$, so it should be $L = \{A + a v : a \in \mathbb R\}$. Then the plane itself is formed by adding in $B$, so the plane is $H = \{A + av + bB : a,b \in \mathbb R\}$. This looks like the plane through the origin $\mathrm{span}\{v, B\}$ shifted by $A$, so I can describe the plane by getting a normal vector $\mathbf n$ to $\mathrm{span}\{v, B\}$ and then a vector $x$ is in the plane exactly when $\mathbf n^T(x - A) = 0$. I found $\mathbf n$ by picking something parallel to $v \times B$ (I got $\mathbf n = (2,1,1)$). But when I do this I get $2x + y + z = 9$ as the equation which my book says is wrong. But I can't see a flaw in my argument. Am I wrong about $H$ or $L$ or something else?
 A: The plane $H$ may not be parallel to the vector $B$;
it is parallel to the vector $v$ and the vector from point $A$ to point $B$, which is $(-5,1,0)$.
Now can you use the cross-product to find a normal vector to it
and get the equation your book says for the plane?
A: The vector $\mathbf{n}=(a,b,c)$ is normal to the plane.  The plane contains a line parallel to $v$, and contains a line parallel to $B-A$.  Let $\mathbf{r}=(x,y,z)$ and $B=(x_0,y_0,z_0)$ be the point on the plane which is given:
So solve the system:
$$\mathbf{n} \cdot \mathbf{v} =0$$
$$\mathbf{n} \cdot (B-A) = 0$$
The equation of the plane is then
$$\mathbf{n}\cdot (\mathbf{r}-B)=0$$
or
$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0.$$
A: If you have another point along the line, like $\boldsymbol{C} = \boldsymbol{A} + \boldsymbol{v} = \pmatrix{1 & 3 & 4}$ then you can define the plane from three points.

*

*Vector normal to plane $$\boldsymbol{n} = (\boldsymbol{A}\times\boldsymbol{B}) + (\boldsymbol{B}\times\boldsymbol{C})+(\boldsymbol{C}\times\boldsymbol{A}) = \pmatrix{3 \\ 15 \\ -3} $$


*Equation of plane based on an arbitrary point on the plane $\boldsymbol{P} = \pmatrix{x & y & z}$ and a known point $A$ (also on the plane). $$ \boldsymbol{n}\cdot \boldsymbol{P} = \boldsymbol{n} \cdot \boldsymbol{A}$$ $$3x+15y-3z=36$$

You can also define the normal by
$$ \boldsymbol{n} = ( \boldsymbol{B}-\boldsymbol{A} ) \times \boldsymbol{v} = \pmatrix{3 \\ 15 \\-3} $$ and proceed with the equation as above.

Note that $(B-A)\times(C-A) = (B-A)\times C - (B-A) \times A = B \times C - A \times C - B \times A = A\times B+B\times C + C \times A$
