My solution as follow Normal(N) did not fit and take abbreviation like(A,B,C) to more simple showing. I wonder whether be true.
$ $$\bf {P_1(x_1,y_1,z_1), P_2(x_2,y_2,z_2), P_3(x_3,y_3,z_3)}$
$$U=\overrightarrow{P_1P_2}$$ $$V=\overrightarrow{P_1P_3}$$
$$ UxV=\det\begin{pmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 &z_3-z_1 \end{pmatrix}$$
$\bf N=((y_2-y_1)(z_3-z_1)-(z_2-z_1)(y_3y_1))[A]i+((x_2-x_1)(z_3-z_1)-(z_2-z_1)(x_3x_1))[B]j+((x_2-x_1)(y_3-y_1)-(y_2-y_1)(x_3x_1))[C]k $
$$ \bf A(x-x_1)+B(y-y_1)+C(z-z_1)+D=0$$
Explicitly, the plane $\mathcal{P}$ through the point $P_0 = (x_0, y_0, z_0)$ is uniquely determined (up to a scalar multiple) by a normal vector $\mathbf n = \langle a,b,c \rangle$ according to the following: a point $P$ lies on $\mathcal{P}$ if and only if $\mathbf n$ and $\overrightarrow{P_0 P}$ are orthogonal if and only if $n \cdot \overrightarrow{P_0 P} = 0$ if and only if $$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0.$$ By setting $d = ax_0 + by_0 + cz_0,$ we have $ax + by + cz = d.$
Given three points $P = (x_0, y_0, z_0),$ $Q,$ and $R,$ the equation of the plane through $P,$ $Q,$ and $R$ can be determined by setting $\mathbf n = \overrightarrow{PQ} \times \overrightarrow{PR}$ and computing the dot product $$0 = \mathbf n \cdot \langle x - x_0, y- y_0, z - z_0 \rangle.$$