Giả sử $A(a,0,0),B(0,b,0),C(0,0,c)$.Phương trình theo đoạn chắn của $(ABC)$ là: $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$Vì $\left\{ \begin{array}{l} M(-4;-9;12)\in(ABC)\\OC=OA+OB\\\frac{4}{OC}=\frac{1}{OA}+\frac{1}{OB} \end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} \frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\\|c|=|a|+|b| \\\frac{4}{|c|}=\frac{1}{|a|}+\frac{1}{|b|}\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} \frac{4}{|a|+|b|}=\frac{|a|+|b|}{|ab|}\\|c|=|a|+|b|
\\\frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} (|a|+|b|)^2=4|ab|\\|c|=|a|+|b|
\\\frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} |a|=|b|\\|c|=|a|+|b|
\\\frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\end{array} \right.\Leftrightarrow \left[ \begin{array}{l} a=-7;b=-7;c=-14\\a=-19;b=-19;c=38\\a=-1;b=1;c=2\\a=11;b=-11;c=22 \end{array} \right.$Khi đó, phương trình của $(ABC)$ là: $\left[ \begin{array}{l} \frac{x}{-7}+\frac{y}{-7}+\frac{z}{-14}=1\Leftrightarrow 2x+2y+z+14=0\\\frac{x}{-19}+\frac{y}{-19}+\frac{z}{38}=1\Leftrightarrow 2x+2y-z+38=0\\\frac{x}{-1}+\frac{y}{1}+\frac{z}{2}=1\Leftrightarrow 2x-2y-z+2=0\\\frac{x}{11}+\frac{y}{-11}+\frac{z}{22}=1\Leftrightarrow 2x-2y+z-22=0 \end{array} \right.$
Giả sử $A(a,0,0),B(0,b,0),C(0,0,c)$.Phương trình theo đoạn chắn của $(ABC)$ là: $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$Vì $\left\{ \begin{array}{l} M(-4;-9;12)\in(ABC)\\OC=OA+OB\\\frac{4}{OC}=\frac{1}{OA}+\frac{1}{OB} \end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} \frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\\c=a+b \\\frac{4}{c}=\frac{1}{a}+\frac{1}{b}\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} \frac{4}{a+b}=\frac{a+b}{ab}\\c=a+b
\\\frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} (a+b)^2=4ab\\c=a+b
\\\frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} a=b\\c=a+b
\\\frac{-4}{a}+\frac{-9}{a}+\frac{12}{2a}=1\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} a=-7\\b=-7\\c=-14 \end{array} \right.$Khi đó, phương trình của $(ABC)$ là: $\frac{x}{-7}+\frac{y}{-7}+\frac{z}{-14}=1\Leftrightarrow 2x+2y+z+14=0$
Giả sử $A(a,0,0),B(0,b,0),C(0,0,c)$.Phương trình theo đoạn chắn của $(ABC)$ là: $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$Vì $\left\{ \begin{array}{l} M(-4;-9;12)\in(ABC)\\OC=OA+OB\\\frac{4}{OC}=\frac{1}{OA}+\frac{1}{OB} \end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} \frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\\
|c
|=
|a
|+
|b
| \\\frac{4}{
|c
|}=\frac{1}{
|a
|}+\frac{1}{
|b
|}\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} \frac{4}{
|a
|+
|b
|}=\frac{
|a
|+
|b
|}{
|ab
|}\\
|c
|=
|a
|+
|b
|
\\\frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l} (
|a
|+
|b
|)^2=4
|ab
|\\
|c
|=
|a
|+
|b
|
\\\frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l}
|a
|=
|b
|\\
|c
|=
|a
|+
|b
|
\\\frac{-4}{a}+\frac{-9}{
b}+\frac{12}{
c}=1\end{array} \right.\Leftrightarrow \left
[ \begin{array}{l} a=-7
;b=-7;c=-14\\
a=-19;b=-
19;c=38\\
a=-1;b=1;c=
2\\a=11;b=-1
1;c=22 \end{array} \right.$Khi đó, phương trình của $(ABC)$ là: $\
left[ \begin{array}{l} \frac{x}{-7}+\frac{y}{-7}+\frac{z}{-14}=1\Leftrightarrow 2x+2y+z+14=0
\\\frac{x}{-19}+\frac{y}{-19}+\frac{z}{38}=1\Leftrightarrow 2x+2y-z+38=0\\\frac{x}{-1}+\frac{y}{1}+\frac{z}{2}=1\Leftrightarrow 2x-2y-z+2=0\\\frac{x}{11}+\frac{y}{-11}+\frac{z}{22}=1\Leftrightarrow 2x-2y+z-22=0 \end{array} \right.$