Find a general equation of a plane passing through points
$P_1\color{green}{\textbf{(1,2,3)}}$, $P_2(4,3,2)$ and $P_3(-3,1,7)$.
Solution

Let us construct two vectors originating from point $P_1$:
$\vec{P_1P_2} = [\ 4-1\ ,\ 3-2\ ,\ 2-3\ ] = [3,1,-1]$,
$\vec{P_1P_3} = [\ -3-1\ ,\ 1-2\ ,\ 7-3\ ] = [-4,-1,4]$,

Then, vector $\vec n$ can be constructed as their cross product:
$\vec{n} = \vec{P_1P_2} \times \vec{P_1P_3}$

$\vec{n} = \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k}\\3 & 1 & -1 \\ -4 & -1 & 4\end{array}\right|$

$\vec{n} = \left[\ \left|\begin{array}{cc} 1 & -1 \\ -1 & 4\end{array}\right|\ ,\ - \left|\begin{array}{cc} 3 & -1 \\ -4 & 4 \end{array}\right|\ ,\ \left|\begin{array}{cc} 3 & 1\\ -4 & -1 \end{array}\right| \ \right]$

$\vec{n} = \color{Blue}{\textbf{[3,-8,1]}}$

Finally, we can write down the equation of plane $\pi$ as:
$\pi: \color{Blue}{\textbf{3}}(x-\color{Green}{\textbf{1}}) \color{Blue}{\textbf{-8}}(y-\color{Green}{\textbf{2}})+ \color{Blue}{\textbf{1}}(z-\color{Green}{\textbf{3}}) = 0$
$\pi: 3x-8y+z+10= 0$

Answer

$\pi: 3x-8y+z+10= 0$.