Proof of well-known 'Intersection Of Three Planes' formula.
Introduction & Motivation
Recently, I found out well-known formula 'Intersection between Three Planes' (Equation (22.31) of [1]) has no proof on literaly anywhere. Even (almost) the book where it is propsed[2].
So, I decided to figure out how this formula is derived.. (of course, experimentally it is wokrs very well.)
Notation
Let there be three planes on euclidean space $P_1, P_2, P_3$ which intersect in a point. If we let there normals as $n_1=(a_1, b_1, c_1), n_2=(a_2, b_2, c_2), n_3=(a_3, b_3, c_3)$, they must be linearly independent.
Also, let the planes $P_1, P_2, P_3$ as the system of equation like below :
$$\begin{cases} a_1x+b_1y+c_1z+d_1=0 \\ a_2x+b_2y+c_2z+d_2=0 \\ a_3x+b_3y+c_3z +d_3=0 \end{cases}$$Proof
$$DET=\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix}$$
Since that, three planes are met in a point (linearly independent)
$$DET \neq 0$$
By Cramer's Rule,
$$(-x)=\frac{\begin{vmatrix} \color{red}{d_1} & b_1 & c_1 \\ \color{red}{d_2} & b_2 & c_2 \\ \color{red}{d_3} & b_3 & c_3 \\ \end{vmatrix}}{DET}, (-y)=\frac{\begin{vmatrix} a_1 & \color{red}{d_1} & c_1 \\ a_2 & \color{red}{d_2} & c_2 \\ a_3 & \color{red}{d_3} & c_3 \\ \end{vmatrix}}{DET}, (-z)=\frac{\begin{vmatrix} a_1 & b_1 & \color{red}{d_1} \\ a_2 & b_2 & \color{red}{d_2} \\ a_3 & b_3 & \color{red}{d_3} \\ \end{vmatrix}}{DET}$$
By cofactor expansion,
$$(-x)=\frac{1}{DET} \left\lbrace d_1\begin{vmatrix} b2 & c2 \\ b3 & c3 \end{vmatrix} -b_1\begin{vmatrix} d2 & c2 \\ d3 & c3 \end{vmatrix} + c_1\begin{vmatrix} d2 & b2 \\ d3 & b3 \end{vmatrix} \right\rbrace \\ = \frac{1}{DET} \left\lbrace \color{purple}{A} -b_1\begin{vmatrix} d2 & c2 \\ d3 & c3 \end{vmatrix} + c_1\begin{vmatrix} d2 & b2 \\ d3 & b3 \end{vmatrix} \right\rbrace $$
$$(-y)=\frac{1}{DET} \left\lbrace a_1\begin{vmatrix} d2 & c2 \\ d3 & c3 \end{vmatrix} -d_1\begin{vmatrix} a2 & c2 \\ a3 & c3 \end{vmatrix} + c_1\begin{vmatrix} a2 & d2 \\ a3 & d3 \end{vmatrix} \right\rbrace \\ = \frac{1}{DET} \left\lbrace a_1\begin{vmatrix} d2 & c2 \\ d3 & c3 \end{vmatrix} + \color{purple}{B} + c_1\begin{vmatrix} a2 & d2 \\ a3 & d3 \end{vmatrix} \right\rbrace $$
$$(-z)=\frac{1}{DET} \left\lbrace a_1\begin{vmatrix} b2 & d2 \\ b3 & d3 \end{vmatrix} -b_1\begin{vmatrix} a2 & d2 \\ a3 & d3 \end{vmatrix} + d_1\begin{vmatrix} a2 & b2 \\ a3 & b3 \end{vmatrix} \right\rbrace \\ = \frac{1}{DET} \left\lbrace a_1\begin{vmatrix} b2 & d2 \\ b3 & d3 \end{vmatrix} -b_1\begin{vmatrix} a2 & d2 \\ a3 & d3 \end{vmatrix} + \color{purple}{C} \right\rbrace $$
By unwind 2x2 deterimants :
$$(-x) = \frac{1}{DET} \left\lbrace \color{purple}{A} - b_1(d_2c_3 - c_2d_3) + c_1(d_2b_3 - b_2d_3) \right\rbrace \\ = \frac{1}{DET} \left\lbrace \color{purple}{A} + \color{blue}{D} + \color{blue}{E} + \color{blue}{F} + \color{blue}{G} \right\rbrace $$
$$(-y) = \frac{1}{DET} \left\lbrace a_1(d_2c_3 - c_2d_3) + \color{purple}{B} - c_1(a_2d_3 - d_2a_3) \right\rbrace \\ = \frac{1}{DET} \left\lbrace \color{blue}{U} + \color{blue}{R} + \color{purple}{B} + \color{blue}{Q} + \color{blue}{H} \right\rbrace $$
$$(-z) = \frac{1}{DET} \left\lbrace a_1(b_2d_3 - d_2b_3) - b_1(a_2d_3 - d_2a_3) + \color{purple}{C} \right\rbrace \\ = \frac{1}{DET} \left\lbrace \color{blue}{L} + \color{blue}{I} + \color{blue}{M} + \color{blue}{K} + \color{purple}{C} \right\rbrace $$
B. Unwind the magic intersection formula.
Below is the formula for calculating a intersection point between three planes (from Equation (22.31) of [1])
$$p_{intersection}=\frac{(p_1 \cdot n_1)(n_2 \times n_3) + (p_2 \cdot n_2)(n_3 \times n_1) + (p_3 \cdot n_3)(n_1 \times n_2)}{\begin{vmatrix} n_1 \\ n_2 \\ n_3 \end{vmatrix} }$$
Since that $p_1, p_2, p_3$ are the point on the plane $P_1, P_2, P_3$ repectively, above can be expressed as :
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