I like this plane

Let the points are $\text{P(1,2,3) , Q(1,-3,2)}$

Then the vector between the points $\vec{a} = (1-1,-3-2,2-3) = (0,-5,-1)$ will lie in the plane.

Direction vector for the z-axis is $\vec{b}=(0,0,1)$ which also lies in the plane.

So a normal vector for the plane is :

$\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 0&-5&-1\\0&0&1\end{vmatrix}$

We need to expand along the third row as all other are 0 , we get

$= \begin{aligned} \widehat{i}\begin{vmatrix} -5&-1\\0&1\end{vmatrix}\end{aligned} = -5\widehat{i}$

Equation of the plane is therefore : - $-5(x-1)=0\implies x=1$

If P = (1,2,3), Q = (1,-3,2). Since this plane is parallel to Z axis, the third point is R=(1,0,0) If equation of the plane is ax+by+cz+d=0, we have 3 equations: a+2b+3c+d=0 a-3b+2c+d=0 a+d=0, which gives d=-a, b=0,c=0. Hence the equation is ax-a=0, OR x=1

let the eq : Ax + By + Cz +D =0 becuse of the plane paralel to z axis so C =0 using (1.2.3) A + 2 B + D =0 (1) using ( 1 , -3 , 2 ) A - 3B +D =0 (2) so B = 0 and a + d = 0 A = - D Ax - A =0 / A x-1 =0 x=1

No need to overcomplicate this one, one of the easier way to do this is to draw the 2 points on xyz plane, project it onto xy plane, make a line through those two projected points in xy plane and you can easily see it forms the line x = 1 (in xy plane), in form to make it parallel just extend it infinitely above or below xy plane so that it forms the plane x = 1. Or just do this on 2D cartesian coordinate plane by throwing away the z components and form a line, it creates the line x = 1 on xy plane, extend it to 3D it becomes the plane x = 1 .