Answer
$7x-y-3z=5$
Work Step by Step
Step 1
Since the plane contains the line with parametric equation
\begin{equation*}
x=-1+t, \quad y=t, \quad z=-4+2t
\end{equation*}
it contains any point on the line. We notice that for $t=0$, $(x,y,z)=(-1,0,-4)$ is on the plane. Also by inspection we notice that $\vec{v}=\langle 1,1,2 \rangle$ is parallel vector to the line, then it is parallel to the plane. Since $\vec{P}$ and $\vec{Q}$ are two point in the plane we can say that the vector $\vec{u}=\vec{PQ}$ is also parallel to the plane that contains these points. Therefore $\vec{n}=\vec{u}\times\vec{v}$ is a normal vector to the plane that satisfies the stated conditions
\begin{align*}
\vec{n} &= \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -3 & 0 & 1 \\ 0 & 1 & 2 \end{vmatrix} = 7\vec{i} -\vec{j} -3\vec{k} = \langle 7,-1,-3\rangle
\end{align*}
Therefore the equation of the plane can be written as
\begin{align*}
7(x-2)-(y-0)-3(z-3)&=0 \\
7x-y-3z&=5
\end{align*}
Finally, the equation of the plane that satisfies the stated conditions can be written as
\begin{equation*}
7x-y-3z=5
\end{equation*}
