How do I align equations along operators within parentheses?

Question

Edit 3: See my own answer if you would like to achieve exactly what I was asking for!

Problem
When I put the alignment character (ampersand) inside parentheses, it breaks the compiler.

% !TEX TS-program = pdflatex
\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{array}
\usepackage{amsmath}
\newcommand{\overbar}[1]{\mkern 1.5mu\overline{\mkern-1.5mu#1\mkern-1.5mu}\mkern 1.5mu}
\newcommand{\minus}{\scalebox{0.6}[1.0]{$-$}}

\begin{document}
    \begin{equation}
    \resizebox{\linewidth}{!}{
        $\begin{aligned}
            \minus\frac{{K_y}_i}{\Delta y^2 \left(1 &+ \frac{K_i^2 \pi^2 \Delta y^2}{12}\right)} \overbar{T}_{i, j-1}\\
            \minus\frac{{K_x}_i}{\Delta x^2\! \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)} \overbar{T}_{i-1, j} + 2 \!\left(\! \frac{{K_x}_i}{\Delta x^2 \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)} \!&+\! \frac{{K_y}_i}{\Delta y^2 \left(1 + \frac{K_i^2 \pi^2 \Delta y^2}{12}\right)} \!\right)\! \overbar{T}_{i, j} - \frac{{K_x}_i}{\Delta x^2\! \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)} \overbar{T}_{i+1, j}\\
            \minus\frac{{K_y}_i}{\Delta y^2 \left(1 &+ \frac{K_i^2 \pi^2 \Delta y^2}{12}\right)} \overbar{T}_{i, j+1}
        \end{aligned}
        =0$
    }
    \end{equation}
\end{document}

Details and Research
I would like to arbitrarily align equations so that I can create a "stencil" representation of a nodal matrix to illustrate the process of FDM assembly. When looking through ~7 posts related to alignment, none of them offered the sort of capability I desire.

What I've Tried
I've achieved a workaround by aligning to a point outside the parenthesis and adding a whole bunch of space:

\begin{equation}
\resizebox{\linewidth}{!}{
    $\begin{aligned}
        &\hspace{3em}\minus\frac{{K_y}_i}{\Delta y^2 \left(1 + \frac{K_i^2 \pi^2 \Delta y^2}{12}\right)} \overbar{T}_{i, j-1}\\
        \minus\frac{{K_x}_i}{\Delta x^2\! \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)} \overbar{T}_{i-1, j} + 2 &\!\left(\! \frac{{K_x}_i}{\Delta x^2 \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)} \!+\! \frac{{K_y}_i}{\Delta y^2 \left(1 + \frac{K_i^2 \pi^2 \Delta y^2}{12}\right)} \!\right)\! \overbar{T}_{i, j} - \frac{{K_x}_i}{\Delta x^2\! \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)} \overbar{T}_{i+1, j}\\
        &\hspace{3em}\minus\frac{{K_y}_i}{\Delta y^2 \left(1 + \frac{K_i^2 \pi^2 \Delta y^2}{12}\right)} \overbar{T}_{i, j+1}
    \end{aligned}
    =0$
}
\end{equation}

I would like a less hacky solution, though.

Thanks so much in advance!

Edit 2: It still hasn't answered my question, but the following improvement will be an easier representation of what I would like to be able to do. I have made the following improvement as per Mico's suggestion. Again, I had to use a workaround, albeit a much cleaner one:

\noindent Let $\kappa_x = \frac{{K_x}_i}{\Delta x^2\! \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)}$ and $\kappa_y = \frac{{K_y}_i}{\Delta y^2 \left(1 + \frac{K_i^2 \pi^2 \Delta y_i^2}{12}\right)}$. Then
\begin{equation}
    \begin{aligned}
        &\minus\kappa_y \overbar{T}_{i, j-1}\\
        \minus\kappa_x \overbar{T}_{i-1, j} + 2 &\left( \kappa_x + \kappa_y \right) \overbar{T}_{i, j} - \kappa_x \overbar{T}_{i+1, j}\\
        &\minus\kappa_y \overbar{T}_{i, j+1}
    \end{aligned}
    =0
\end{equation}

Answer 1

Remark: I thoroughly updated this answer after learning from the OP what the (to me) unusual cruciform layout of the five additive terms across three lines was meant to signify.

The following one-line equation should be easy to grasp for your readers. (Note that I've multiplied the terms on the left hand side by -1, which is permissible since their sum equals 0.) With this setup, do take a sentence or two to explain the ordering of elements in terms of items involving T_{.,j-1}, T{.,j}, and T_{.,j+1.

\documentclass[11pt]{article}
\let\overbar\overline  % ?
\begin{document}

Set $\kappa_x={K_x}_i\big/\bigl[\Delta x^2(1-\pi^2)/12\bigr]$ and 
$\kappa_y={K_y}_i\big/\bigl[\Delta y^2(1+K_i^2\pi^2 )/12\bigr]$. 
Then
\begin{equation}
  \kappa_y \overbar{T}_{i,j-1} 
+ \kappa_x \overbar{T}_{i-1,j} 
- 2(\kappa_x + \kappa_y) \overbar{T}_{i,j} 
+ \kappa_x \overbar{T}_{i+1,j}
+ \kappa_y \overbar{T}_{i,j+1}
=0
\end{equation}

\end{document}

Answer 2

I think that the other options discussed here look much better and are significantly more readable, but if you are using LuaLaTeX and really want to do such odd alignments you can use my luamathalign package.

It introduces an \AlignHere macro which acts like & except that it doesn't have all these pesky limitations about where it can't appear: Just load luamathalign and replace every problematic & with \AlignHere.

% !TEX TS-program = lualatex
\documentclass{article}
\usepackage{unicode-math,luamathalign}
\begin{document}
\begin{equation}
    \begin{aligned}
        \minus\frac{{K_y}_i}{\Delta y^2 \left(1 \AlignHere+ \frac{K_i^2 \pi^2 \Delta y^2}{12}\right)} \overbar{T}_{i, j-1}\\
        \minus\frac{{K_x}_i}{\Delta x^2\! \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)} \overbar{T}_{i-1, j} + 2 \!\left(\! \frac{{K_x}_i}{\Delta x^2 \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)} \!\AlignHere+\! \frac{{K_y}_i}{\Delta y^2 \left(1 + \frac{K_i^2 \pi^2 \Delta y^2}{12}\right)} \!\right)\! \overbar{T}_{i, j} - \frac{{K_x}_i}{\Delta x^2\! \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)} \overbar{T}_{i+1, j}\\
        \minus\frac{{K_y}_i}{\Delta y^2 \left(1 \AlignHere + \frac{K_i^2 \pi^2 \Delta y^2}{12}\right)} \overbar{T}_{i, j+1}
    \end{aligned}
    =0
\end{equation}

\end{document}

Answer 3

Edit 2: See the bonus solution to be able to to exactly what the original question asked!

Welp, it turns out I should have more deeply considered one of the other posts I saw relating to just putting stuff in a matrix. If I do the following, it aligns perfectly and is clearer that the entirety of the terms are equal to zero. Credit to @Mico for leading me to this answer!

\noindent Let $\kappa_x = \frac{{K_x}_i}{\Delta x^2\! \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)}$ and $\kappa_y = \frac{{K_y}_i}{\Delta y^2 \left(1 + \frac{K_i^2 \pi^2 \Delta y_i^2}{12}\right)}$. Then
\begin{equation}
    \begin{pmatrix}
        \minus\kappa_y \overbar{T}_{i-1, j}\\
        \minus\kappa_x \overbar{T}_{i, j-1} + 2 \left( \kappa_x + \kappa_y \right) \overbar{T}_{i, j} - \kappa_x \overbar{T}_{i, j+1}\\
        \minus\kappa_y \overbar{T}_{i+1, j}
    \end{pmatrix}
    =0
\end{equation}

Which results in the following:

Now that I think about it, if I didn't want the parenthesis, I could even do the following:

\noindent Let $\kappa_x = \frac{{K_x}_i}{\Delta x^2\! \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)}$ and $\kappa_y = \frac{{K_y}_i}{\Delta y^2 \left(1 + \frac{K_i^2 \pi^2 \Delta y_i^2}{12}\right)}$. Then
\begin{equation}
    \begin{matrix}
        \minus\kappa_y \overbar{T}_{i-1, j}\\
        \minus\kappa_x \overbar{T}_{i, j-1} + 2 \left( \kappa_x + \kappa_y \right) \overbar{T}_{i, j} - \kappa_x \overbar{T}_{i, j+1}\\
        \minus\kappa_y \overbar{T}_{i+1, j}
    \end{matrix}
    =0
\end{equation}

Which would result in the following:

Edit: Bonus solution! (Which happens to be the answer closest to the original question) In case you do indeed something tall with \left and \right (which I ended up not needing), Just open and close the right and left around the alignment operator and add a \vphantom{<something with the same height as what you want your brackets/braces/parenthesis to be>}:

\noindent Let $\kappa_x = \frac{{K_x}_i}{\Delta x^2\! \left(1 - \frac{\pi^2 \Delta x^2}{12}\right)}$ and $\kappa_y = \frac{{K_y}_i}{\Delta y^2 \left(1 + \frac{K_i^2 \pi^2 \Delta y_i^2}{12}\right)}$. Then
\begin{equation}
    \begin{aligned}
        \minus\kappa_y &\overbar{T}_{i, j-1}\\
        \minus\kappa_x \overbar{T}_{i-1, j} + 2 \left(\kappa_x \vphantom{\kappa_x} \right. &+ \left.\vphantom{\kappa_y} \kappa_y\right) \overbar{T}_{i, j} - \kappa_x \overbar{T}_{i+1, j}\\
        \minus\kappa_y &\overbar{T}_{i, j+1}
    \end{aligned}
    =0
\end{equation}