The normal equation of the plane that contains the line $(1,1,1) + t(-2,0,3)$ 
Determine the equation of the plane that contains the point $(4,2,-1)$
  and also the line $L: (1,1,1) + t(-2,0,3)$ for $t\in\mathbb{R}$.

The direction vector $(-2,0,3)$ of the line is also a direction vector for the plane.
We can get a vector orthogonal to the plane by solving
$$(x,y,z) \cdot (-2,0,3) = 0$$
$$-2x+3z = 0$$
One valid solution would be $(3,1,2)$.
So we have that $(3,1,2)$ is orthogonal to the plane. And since $(4,2,-1)$ belongs to it, a normal equation for the plane can be calculated this way:
$$(x,y,z)\cdot (3,1,2) = (4,2,-1) \cdot (3,1,2)$$
$$3x+y+2z = 12$$

But that's wrong. If the plane contains the line $L$, it should also contain the point
$$(1,1,1)+5\cdot(-2,0,3) = (-9,1,16)$$
But it doesn't:
$$3(-9)+(1)+2(16) = -27 + 1 + 32 = 8 \not = 12$$
What did I do wrong?
 A: One valid solution would be $(3,1,2)$.
This is incorrect. Not all vector perpendicular to $(-2,0,3)$ gives you the normal vector of the plane. The correct way is pick a point on the line, for simplicity say $(1,1,1)$ then the normal vector of plane is perpendicular to $(-2,0,3)$ and $(4,2,-1)-(1,1,1)=(3,1,-2)$. Hence you need to calculate $n=(-2,0,3)\times (3,1,-2)$
Then the plane is given by $<(x,y,z)-(1,1,1), n>=0$
A: An easier way is: 
You already have one vector and one point then:
Let $L : B + t\overrightarrow{w}$ where $\overrightarrow{w} = (-2,0,3) \wedge B = (1,1,1), t \in \mathcal{R}$ 
Let $ C = (4,2,-1)$ 
You look for a plane that contains this line and $C$, now:
Let $\overrightarrow{z} = \overrightarrow{CB} = \overrightarrow{B-C}$ Let $H$ be the plane defined by the parametric equation:
$H : B \ + \ h \overrightarrow{z} \ + \ t \overrightarrow{w}; t,h \in \mathcal{R}$
This plane portrays these conditions(Portrays? I'm sorry for my English) 
Another approach could be:
Let $\overrightarrow{a} = \overrightarrow{w} \times \overrightarrow{z}$ then it's $(x,y,z)$ coordinates represent the coefficients of the Cartesian equation of the desired plane. Let these coordinates be $a_1,a_2 \wedge a_3$ respectively: now the equation of the plane:
$a_1(x-4) + a_2(y-2) + a_3(z+1)= 0$ is the Cartesian equation of the plane.
A: The plane is passing through two points and one line....
$(x_1,y_1,z_1) = (4,2,−1)$
$(x_2,y_2,z_2) = (1,1,1)$
$(l,m,n) = (−2,0,3)$
Cartesian equation
\begin{equation}
\begin{vmatrix}
x-x_1 & y-y_1 & z-z_1\\
x_2-x_1 &y_2-y_1 &z_2-z_1\\
l    & m    & n 
\end{vmatrix} = 0
\end{equation}
