Answer
$4x+13y-z=1$
Work Step by Step
Step 1
Since the plane passes through $\mathbf{P}_1=(-2,1,4)$ and $\mathbf{P}_2=(1,0,3)$, the vector $\overrightarrow{\mathbf{v}}=\mathbf{P}_1\mathbf{P}_2=\begin{pmatrix}1 \ 0 \ 3\end{pmatrix}-\begin{pmatrix}-2 \ 1 \ 4\end{pmatrix}=\begin{pmatrix}3 \ -1 \ -1\end{pmatrix}$ is in the plane.
On the other hand, the normal vector to the given plane $4x-y+3z=2$ is $\mathbf{u}=\begin{pmatrix}4 \ -1 \ 3\end{pmatrix}$, which is also parallel to the plane. Therefore, we have that $\overrightarrow{\mathbf{v}}$ is orthogonal to $\mathbf{u}$.
Hence, we can find a normal vector to the plane that satisfies the stated conditions by taking the cross product of $\overrightarrow{\mathbf{v}}$ and $\mathbf{u}$:
$\mathbf{n}=\overrightarrow{\mathbf{v}}\times \mathbf{u}=\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & -1 \\ 4 & -1 & 3\end{vmatrix}=-4\mathbf{i}-13\mathbf{j}+\mathbf{k}=\begin{pmatrix}-4 \ -13 \ 1\end{pmatrix}$
Also, since the plane passes through $\mathbf{P}_1=(-2,1,4)$, we have that the equation of the plane can be written as:
$-4(x+2)-13(y-1)+(z-4)=0 \implies 4x+13y-z=1$
Therefore, the equation of the plane is:
$4\mathbf{x}+13\mathbf{y}-\mathbf{z}=1$
