# How can I solve a list of problems and get a LaTeX file?

I have a list of problems

lis = {{{0, 1, 5}, {7, 4, 2}, {9, 3, 1}}, {{6, 8, 5}, {7, 4, 3}, {9,
1, 2}}, {{6, 8, 5}, {7, 4, 3}, {9, 2, 1}}, {{6, 9, 1}, {7, 2,
3}, {8, 4, 5}}, {{6, 9, 1}, {7, 2, 3}, {8, 5, 4}}, {{6, 9,
1}, {7, 2, 4}, {8, 3, 5}}};
data = Table[{pA, pB, pC} = points;
mylist = {pA, pB, pC,
Subtract @@ Expand[CoplanarPoints[{pA, pB, pC, {x, y, z}}]]] ==
0}, {points, lis}];


and I want to write solutions of all problems.

SetDirectory[NotebookDirectory[]]
lis = {{{0, 1, 5}, {7, 4, 2}, {9, 3, 1}}, {{6, 8, 5}, {7, 4, 3}, {9,
1, 2}}, {{6, 8, 5}, {7, 4, 3}, {9, 2, 1}}, {{6, 9, 1}, {7, 2,
3}, {8, 4, 5}}, {{6, 9, 1}, {7, 2, 3}, {8, 5, 4}}, {{6, 9,
1}, {7, 2, 4}, {8, 3, 5}}};
data = Table[{pA, pB, pC} = points;
mylist = {pA, pB, pC,
Subtract @@ Expand[CoplanarPoints[{pA, pB, pC, {x, y, z}}]]] ==
0}, {points, lis}];
toX[e_] :=
StringReplace[ToString[TeXForm[e]], {"\\}" -> ")", "\\{" -> "("}]
fileName = FileNameJoin[{Directory[], "my_HW.tex"}]
If[FileExistsQ[fileName], DeleteFile[fileName]];
file = OpenWrite[fileName, PageWidth -> Infinity];
WriteString[file,
"\\documentclass[12pt,a4paper]{article}\n" <>
"\\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}\n\
" <> "\\usepackage{amsmath}\n" <> "\\usepackage{amsthm}\n" <>
"\\usepackage{esvect}\n" <> "\\theoremstyle{definition}\n" <>
"\\newtheorem{ex}{Exercise}\n" <> "\\newtheorem{sol}{Solution}\n" <>
"\\begin{document}\n"];
Do[WriteString[file,
"\\begin{ex}\n" <>
"The equation of the plane passing through three points $$A" <> toX[data[[n, 1]]] <> "$$, $$B" <> toX[data[[n, 2]]] <> "$$, $$C" <> toX[data[[n, 3]]] <> "$$ is $$" <> toX[data[[n, 4]]] <> ".$$\n\\end{ex}\n
\\begin{sol}
We have
$$\vv{AB} == data[[n,2]] - data[[n,1]]$$;
$$\vv{AC}==data[[n,3]] - data[[n,2]]$$;
Noralvector of the plane $$(ABC)$$ is $$\vv{n} = Cross[AB,AC]$$.
The equation of the plane $$(ABC)$$ is
\\end{sol}
"], {n, 1, Length@data}];
WriteString[file, "\\end{document}\n"];
Close[file]


My LaTeX file

    \documentclass[12pt,a4paper]{article}
\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{esvect}
\theoremstyle{definition}
\newtheorem{ex}{Exercise}
\newtheorem{sol}{Solution}
\begin{document}
\begin{ex}
The equation of the plane passing through three points $A(0,1,5)$, $B(7,4,2)$, $C(9,3,1)$ is $6 x-y+13 z-64=0.$
\end{ex}

$$\begin{sol} We have \vv{AB} == data[[n,2]] - data[[n,1]]; \vv{AC}==data[[n,3]] - data[[n,2]]; Noralvector of the plane (ABC) is \vv{n} = Cross[AB,AC]. The equation of the plane (ABC) is \end{sol}$$
$$\begin{ex} The equation of the plane passing through three points A(6,8,5), B(7,4,3), C(9,1,2) is 2 x+3 y-5 z-11=0. \end{ex}$$

$$\begin{sol} We have \vv{AB} == data[[n,2]] - data[[n,1]]; \vv{AC}==data[[n,3]] - data[[n,2]]; Noralvector of the plane (ABC) is \vv{n} = Cross[AB,AC]. The equation of the plane (ABC) is \end{sol}$$
$$\begin{ex} The equation of the plane passing through three points A(6,8,5), B(7,4,3), C(9,2,1) is 2 x-y+3 z-19=0. \end{ex}$$

$$\begin{sol} We have \vv{AB} == data[[n,2]] - data[[n,1]]; \vv{AC}==data[[n,3]] - data[[n,2]]; Noralvector of the plane (ABC) is \vv{n} = Cross[AB,AC]. The equation of the plane (ABC) is \end{sol}$$
$$\begin{ex} The equation of the plane passing through three points A(6,9,1), B(7,2,3), C(8,4,5) is 2 x-z-11=0. \end{ex}$$

$$\begin{sol} We have \vv{AB} == data[[n,2]] - data[[n,1]]; \vv{AC}==data[[n,3]] - data[[n,2]]; Noralvector of the plane (ABC) is \vv{n} = Cross[AB,AC]. The equation of the plane (ABC) is \end{sol}$$
$$\begin{ex} The equation of the plane passing through three points A(6,9,1), B(7,2,3), C(8,5,4) is 13 x-y-10 z-59=0. \end{ex}$$

$$\begin{sol} We have \vv{AB} == data[[n,2]] - data[[n,1]]; \vv{AC}==data[[n,3]] - data[[n,2]]; Noralvector of the plane (ABC) is \vv{n} = Cross[AB,AC]. The equation of the plane (ABC) is \end{sol}$$
$$\begin{ex} The equation of the plane passing through three points A(6,9,1), B(7,2,4), C(8,3,5) is 5 x-y-4 z-17=0. \end{ex}$$

$$\begin{sol} We have \vv{AB} == data[[n,2]] - data[[n,1]]; \vv{AC}==data[[n,3]] - data[[n,2]]; Noralvector of the plane (ABC) is \vv{n} = Cross[AB,AC]. The equation of the plane (ABC) is \end{sol}$$
\end{document}


I got

My LaTeX based on

Table[{ab = data[[n, 2]] - data[[n, 1]],
ac = data[[n, 3]] - data[[n, 1]], n = Cross[ab, ac]}, {n, 1, Length@data}]


It looks like some of your solutions are wrong. At least this is what my code shows. For example problem 3 gives different solution that what you have. You might want double checks how to you obtained the equations of the plane. I used standard method in text book.

Since I wrote this in code cells, it will be messed up when posting. So I include also links to notebook,PDF and latex

PDF

LATEX

NOTEBOOK

Here is screen shot of first page of the PDF

## Mathematica code

SetDirectory[NotebookDirectory[]]
lis={{{0,1,5},{7,4,2},{9,3,1}},{{6,8,5},{7,4,3},{9,1,2}},{{6,8,5},{7,4,3},{9,2,1}},{{6,9,1},{7,2,3},{8,4,5}},{{6,9,1},{7,2,3},{8,5,4}},{{6,9,1},{7,2,4},{8,3,5}}};
data = Table[{pA, pB, pC} = points;
mylist = {pA, pB, pC,
Subtract @@ Expand[CoplanarPoints[{pA, pB, pC, {x, y, z}}]]] ==
0}, {points, lis}];
fileName=FileNameJoin[{Directory[],"my_HW.tex"}];

toX[e_] :=ToString[TeXForm[e]]
fix[s_String]:=StringReplace[s,{"\\}" -> ")", "\\{" -> "("}]

solvePlaneEquation[p1_List,p2_List,p3_List,sol_,x_Symbol,y_Symbol,z_Symbol]:=Module[{n,v1,v2,a,b,c,x0,y0,z0,eq,s},
s="\\begin{ex}\n" <>
"The equation of the plane passing through three points $$A"<> fix[toX[p1]] <> "$$, $$B" <> fix[toX[p2]] <> "$$, $$C" <> fix[toX[p3]] <> "$$ is $$" <> toX[sol] <> "$$\n\\end{ex}\n"<>
"\\begin{sol}\n";
v1=p2-p1;
v2=p3-p1;
n=makeCrossProduct[v1,v2];
{a,b,c}=n;
{x0,y0,z0}=p1;
eq=Expand[a*(x-x0)+b*(y-y0)+c*(z-z0)]==0;
(*below ref cvgmt from https://mathematica.stackexchange.com/questions/290284/move-variables-to-one-side-of-equation*)
(*move constant to rhs to make it better looking*)
Print["eq=",eq];
eq=SubtractSides[SubtractSides[eq],First@CoefficientArrays[eq]];

s=s<>"We have\n\\begin{align*}\n\\overrightarrow{AB}&="<>fix@toX[p2]<>"-"<>fix@toX[p1]<>"\\\\ \n"<>
"&="<>fix@toX[v1]<>"\\\\ \n"<>
"\\overrightarrow{AC}&="<>fix@toX[p3]<>"-"<>fix@toX[p1]<>"\\\\ \n"<>
"&="<>fix@toX[v2]<>"\n"<>
"\\end{align*}\n"<>
"Therefore the normal vector $$\\vec{n}$$ to the plane is given by\n"<>
"\\begin{align*}\n"<>
"\\vec{n} &= \\overrightarrow{AB} \\times \\overrightarrow{AC}\\\\ \n"<>
"&="<>fix@toX[v1]<>"\\times"<>fix@toX[v2]<>"\\\\ \n"<>
"&="<>fix@toX[n]<>"\n"<>
"\\end{align*}\n"<>
"Assigning $$a,b,c$$ to the coordinates of the normal vector $$\\vec{n}$$ and "<>
"Assigning $$x_0,y_0,z_0$$ to the coordinates of the vector $$A$$, then the "<>
"equation of the plane is given by\n"<>
"\\begin{align*}\n"<>
"a(x-x_0)+b (y-y_0)+c (z-z_0) &=0\n"<>
"\\end{align*}\n"<>
"Which results in\n"<>
"\\begin{align*}\n"<>
toX[a]<>"\\left(x"<>If[x0>=0,"-","+"]<>toX[x0]<>"\\right)"<>If[b>=0,"+",""]<>
toX[b]<>"\\left(y"<>If[y0>=0,"-","+"]<>toX[y0]<>"\\right)"<>If[c>=0,"+",""]<>
toX[c]<>"\\left(z"<>If[z0>=0,"-","+"]<>toX[z0]<>"\\right)&=0\\\\ \n"<>
toX[eq[[1]]]<>"&="<>toX[eq[[2]]]<>"\n"<>
"\\end{align*}\n"<>
"\\end{sol}\n";

s
]

makeCrossProduct[v1_List,v2_List]:=Module[{},
Cross[v1,v2]
]

processList[L_List,fileName_String,x_Symbol,y_Symbol,z_Symbol]:=Module[{s,file,p1,p2,p3,eq,n},
If[FileExistsQ[fileName],DeleteFile[fileName]];
file=OpenWrite[fileName,PageWidth->Infinity];
WriteString[file,
"\\documentclass[12pt,a4paper]{article}\n" <>
"\\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}\n"<>
"\\usepackage{amsmath}\n" <>
"\\usepackage{amsthm}\n" <>
"\\usepackage{esvect}\n" <>
"\\theoremstyle{definition}\n" <>
"\\newtheorem{ex}{Exercise}\n" <>
"\\newtheorem{sol}{Solution}\n" <>
"\\begin{document}\n"];
Do[p1=L[[n,1]];p2=L[[n,2]];p3=L[[n,3]];eq=L[[n,4]];
s=solvePlaneEquation[p1,p2,p3,eq,x,y,z];
WriteString[file,s]
,{n,1,Length[L]}
];
WriteString[file, "\\end{document}\n"];
Close[file]
]


To run and generate the Latex file

    processList[data, fileName, x, y, z]


## Latex generated by Mathematica

\documentclass[12pt,a4paper]{article}
\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{esvect}
\theoremstyle{definition}
\newtheorem{ex}{Exercise}
\newtheorem{sol}{Solution}
\begin{document}
\begin{ex}
The equation of the plane passing through three points $A(0,1,5)$, $B(7,4,2)$, $C(9,3,1)$ is $6 x-y+13 z-64=0$
\end{ex}
\begin{sol}
We have
\begin{align*}
\overrightarrow{AB}&=(7,4,2)-(0,1,5)\\
&=(7,3,-3)\\
\overrightarrow{AC}&=(9,3,1)-(0,1,5)\\
&=(9,2,-4)
\end{align*}
Therefore the normal vector $\vec{n}$ to the plane is given by
\begin{align*}
\vec{n} &= \overrightarrow{AB} \times \overrightarrow{AC}\\
&=(7,3,-3)\times(9,2,-4)\\
&=(-6,1,-13)
\end{align*}
Assigning $a,b,c$ to the coordinates of the normal vector $\vec{n}$ and Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the equation of the plane is given by
\begin{align*}
a(x-x_0)+b (y-y_0)+c (z-z_0) &=0
\end{align*}
Which results in
\begin{align*}
-6\left(x-0\right)+1\left(y-1\right)-13\left(z-5\right)&=0\\
-6 x+y-13 z&=-64
\end{align*}
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,8,5)$, $B(7,4,3)$, $C(9,1,2)$ is $2 x+3 y-5 z-11=0$
\end{ex}
\begin{sol}
We have
\begin{align*}
\overrightarrow{AB}&=(7,4,3)-(6,8,5)\\
&=(1,-4,-2)\\
\overrightarrow{AC}&=(9,1,2)-(6,8,5)\\
&=(3,-7,-3)
\end{align*}
Therefore the normal vector $\vec{n}$ to the plane is given by
\begin{align*}
\vec{n} &= \overrightarrow{AB} \times \overrightarrow{AC}\\
&=(1,-4,-2)\times(3,-7,-3)\\
&=(-2,-3,5)
\end{align*}
Assigning $a,b,c$ to the coordinates of the normal vector $\vec{n}$ and Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the equation of the plane is given by
\begin{align*}
a(x-x_0)+b (y-y_0)+c (z-z_0) &=0
\end{align*}
Which results in
\begin{align*}
-2\left(x-6\right)-3\left(y-8\right)+5\left(z-5\right)&=0\\
-2 x-3 y+5 z&=-11
\end{align*}
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,8,5)$, $B(7,4,3)$, $C(9,2,1)$ is $2 x-y+3 z-19=0$
\end{ex}
\begin{sol}
We have
\begin{align*}
\overrightarrow{AB}&=(7,4,3)-(6,8,5)\\
&=(1,-4,-2)\\
\overrightarrow{AC}&=(9,2,1)-(6,8,5)\\
&=(3,-6,-4)
\end{align*}
Therefore the normal vector $\vec{n}$ to the plane is given by
\begin{align*}
\vec{n} &= \overrightarrow{AB} \times \overrightarrow{AC}\\
&=(1,-4,-2)\times(3,-6,-4)\\
&=(4,-2,6)
\end{align*}
Assigning $a,b,c$ to the coordinates of the normal vector $\vec{n}$ and Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the equation of the plane is given by
\begin{align*}
a(x-x_0)+b (y-y_0)+c (z-z_0) &=0
\end{align*}
Which results in
\begin{align*}
4\left(x-6\right)-2\left(y-8\right)+6\left(z-5\right)&=0\\
4 x-2 y+6 z&=38
\end{align*}
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,9,1)$, $B(7,2,3)$, $C(8,4,5)$ is $2 x-z-11=0$
\end{ex}
\begin{sol}
We have
\begin{align*}
\overrightarrow{AB}&=(7,2,3)-(6,9,1)\\
&=(1,-7,2)\\
\overrightarrow{AC}&=(8,4,5)-(6,9,1)\\
&=(2,-5,4)
\end{align*}
Therefore the normal vector $\vec{n}$ to the plane is given by
\begin{align*}
\vec{n} &= \overrightarrow{AB} \times \overrightarrow{AC}\\
&=(1,-7,2)\times(2,-5,4)\\
&=(-18,0,9)
\end{align*}
Assigning $a,b,c$ to the coordinates of the normal vector $\vec{n}$ and Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the equation of the plane is given by
\begin{align*}
a(x-x_0)+b (y-y_0)+c (z-z_0) &=0
\end{align*}
Which results in
\begin{align*}
-18\left(x-6\right)+0\left(y-9\right)+9\left(z-1\right)&=0\\
9 z-18 x&=-99
\end{align*}
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,9,1)$, $B(7,2,3)$, $C(8,5,4)$ is $13 x-y-10 z-59=0$
\end{ex}
\begin{sol}
We have
\begin{align*}
\overrightarrow{AB}&=(7,2,3)-(6,9,1)\\
&=(1,-7,2)\\
\overrightarrow{AC}&=(8,5,4)-(6,9,1)\\
&=(2,-4,3)
\end{align*}
Therefore the normal vector $\vec{n}$ to the plane is given by
\begin{align*}
\vec{n} &= \overrightarrow{AB} \times \overrightarrow{AC}\\
&=(1,-7,2)\times(2,-4,3)\\
&=(-13,1,10)
\end{align*}
Assigning $a,b,c$ to the coordinates of the normal vector $\vec{n}$ and Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the equation of the plane is given by
\begin{align*}
a(x-x_0)+b (y-y_0)+c (z-z_0) &=0
\end{align*}
Which results in
\begin{align*}
-13\left(x-6\right)+1\left(y-9\right)+10\left(z-1\right)&=0\\
-13 x+y+10 z&=-59
\end{align*}
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,9,1)$, $B(7,2,4)$, $C(8,3,5)$ is $5 x-y-4 z-17=0$
\end{ex}
\begin{sol}
We have
\begin{align*}
\overrightarrow{AB}&=(7,2,4)-(6,9,1)\\
&=(1,-7,3)\\
\overrightarrow{AC}&=(8,3,5)-(6,9,1)\\
&=(2,-6,4)
\end{align*}
Therefore the normal vector $\vec{n}$ to the plane is given by
\begin{align*}
\vec{n} &= \overrightarrow{AB} \times \overrightarrow{AC}\\
&=(1,-7,3)\times(2,-6,4)\\
&=(-10,2,8)
\end{align*}
Assigning $a,b,c$ to the coordinates of the normal vector $\vec{n}$ and Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the equation of the plane is given by
\begin{align*}
a(x-x_0)+b (y-y_0)+c (z-z_0) &=0
\end{align*}
Which results in
\begin{align*}
-10\left(x-6\right)+2\left(y-9\right)+8\left(z-1\right)&=0\\
-10 x+2 y+8 z&=-34
\end{align*}
\end{sol}
\end{document}

• The result of your code incorrect at "Which results in". Can I use the result of mylist[[4]]. It is correct and was reduced. Commented May 26 at 9:48
• @JohnPaulPeter I think there might be a bug in my code give me time to look more will update if needed. Commented May 26 at 10:10
• I surprise, the results in your PDF file are correct. Only not similart to mylist. Commented May 26 at 10:30
• @JohnPaulPeter I doubled checked and see nothing wrong. Added extra step to make it more clear. So I do not know why your solution for few of them are different. I also updated the links. I thought I copied your list OK. Anyway, if you find bug just let me know and will check again. Commented May 26 at 10:34
• @Nasser Have you ever done a multiple choice in Mathematica and make a latex file? I do know make it. Commented Jun 9 at 3:55

You can try this code

SetDirectory[NotebookDirectory[]]
lis = {{{0, 1, 5}, {7, 4, 2}, {9, 3, 1}}, {{6, 8, 5}, {7, 4, 3}, {9,
1, 2}}, {{6, 8, 5}, {7, 4, 3}, {9, 2, 1}}, {{6, 9, 1}, {7, 2,
3}, {8, 4, 5}}, {{6, 9, 1}, {7, 2, 3}, {8, 5, 4}}, {{6, 9,
1}, {7, 2, 4}, {8, 3, 5}}};
data = Table[{pA, pB, pC} = points;
mylist = {pA, pB, pC, pB - pA, pC - pA,
cr = Cross[pB - pA, pC - pA],
cr[[1]] (x - pA[[1]]) + cr[[2]] (y - pA[[2]]) +
cr[[3]] (z - pA[[3]])] == 0,
Subtract @@ Expand[CoplanarPoints[{pA, pB, pC, {x, y, z}}]]] ==
0}, {points, lis}];
toX[e_] :=
StringReplace[ToString[TeXForm[e]], {"\\}" -> ")", "\\{" -> "("}]
If[FileExistsQ[fileName], DeleteFile[fileName]];
file = OpenWrite[fileName, PageWidth -> Infinity];
WriteString[file,
"\\documentclass[12pt,a4paper]{article}\n" <>
"\\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}\n\
" <> "\\usepackage{amsmath}\n" <> "\\usepackage{amsthm}\n" <>
"\\usepackage{esvect}\n" <> "\\usepackage{fouriernc}\n" <>
"\\theoremstyle{definition}\n" <> "\\newtheorem{ex}{Exercise}\n" <>
"\\newtheorem{sol}{Solution}\n" <> "\\begin{document}\n"];
Do[WriteString[file,
"\\begin{ex}\n" <>
"The equation of the plane passing through three points $$A" <> toX[data[[n, 1]]] <> "$$, $$B" <> toX[data[[n, 2]]] <> "$$, $$C" <> toX[data[[n, 3]]] <> "$$ is $$" <> toX[data[[n, 8]]] <> ".$$\n\\end{ex}\n
\\begin{sol}
\mbox{}
\\begin{itemize}
\item We have
$\vv{AB} = " <> toX[data[[n, 2]]] <> " - " <> toX[data[[n, 1]]] <> "=" <> toX[data[[n, 4]]] <> "$
and
$\vv{AC} = " <> toX[data[[n, 3]]] <> " - " <> toX[data[[n, 1]]] <> "=" <> toX[data[[n, 5]]] <> ".$
\item Therefore the normal vector  to the plane is given by  $$\vv{n}= \ \vv{AB}\\times \vv{AC} = " <> toX[data[[n, 6]]] <> ".$$
\item Equation of the plane is given by   $" <> toX[data[[n, 7]]] <> ".$
Expand and simplify, we get
$$" <> toX[data[[n, 8]]] <> "$$.
\\end{itemize} \qed
\\end{sol}
"], {n, 1, Length@data}];
WriteString[file, "\\end{document}\n"];
Close[file]


The LaTeX file

\documentclass[12pt,a4paper]{article}
\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{esvect}
\usepackage{fouriernc}
\theoremstyle{definition}
\newtheorem{ex}{Exercise}
\newtheorem{sol}{Solution}
\begin{document}
\begin{ex}
The equation of the plane passing through three points $A(0,1,5)$, $B(7,4,2)$, $C(9,3,1)$ is $6 x-y+13 z-64=0.$
\end{ex}

$$\begin{sol} \mbox{} \begin{itemize} \item We have $\vv{AB} = (7,4,2) - (0,1,5)=(7,3,-3)$ and $\vv{AC} = (9,3,1) - (0,1,5)=(9,2,-4).$ \item Therefore the normal vector to the plane is given by \vv{n}= \vv{AB}\times \vv{AC} = (-6,1,-13). \item Equation of the plane is given by $-6 x+y-13 (z-5)-1=0.$ Expand and simplify, we get 6 x-y+13 z-64=0. \end{itemize} \qed \end{sol}$$
$$\begin{ex} The equation of the plane passing through three points A(6,8,5), B(7,4,3), C(9,1,2) is 2 x+3 y-5 z-11=0. \end{ex}$$

$$\begin{sol} \mbox{} \begin{itemize} \item We have $\vv{AB} = (7,4,3) - (6,8,5)=(1,-4,-2)$ and $\vv{AC} = (9,1,2) - (6,8,5)=(3,-7,-3).$ \item Therefore the normal vector to the plane is given by \vv{n}= \vv{AB}\times \vv{AC} = (-2,-3,5). \item Equation of the plane is given by $-2 (x-6)-3 (y-8)+5 (z-5)=0.$ Expand and simplify, we get 2 x+3 y-5 z-11=0. \end{itemize} \qed \end{sol}$$
$$\begin{ex} The equation of the plane passing through three points A(6,8,5), B(7,4,3), C(9,2,1) is 2 x-y+3 z-19=0. \end{ex}$$

$$\begin{sol} \mbox{} \begin{itemize} \item We have $\vv{AB} = (7,4,3) - (6,8,5)=(1,-4,-2)$ and $\vv{AC} = (9,2,1) - (6,8,5)=(3,-6,-4).$ \item Therefore the normal vector to the plane is given by \vv{n}= \vv{AB}\times \vv{AC} = (4,-2,6). \item Equation of the plane is given by $4 (x-6)-2 (y-8)+6 (z-5)=0.$ Expand and simplify, we get 2 x-y+3 z-19=0. \end{itemize} \qed \end{sol}$$
$$\begin{ex} The equation of the plane passing through three points A(6,9,1), B(7,2,3), C(8,4,5) is 2 x-z-11=0. \end{ex}$$

$$\begin{sol} \mbox{} \begin{itemize} \item We have $\vv{AB} = (7,2,3) - (6,9,1)=(1,-7,2)$ and $\vv{AC} = (8,4,5) - (6,9,1)=(2,-5,4).$ \item Therefore the normal vector to the plane is given by \vv{n}= \vv{AB}\times \vv{AC} = (-18,0,9). \item Equation of the plane is given by $9 (z-1)-18 (x-6)=0.$ Expand and simplify, we get 2 x-z-11=0. \end{itemize} \qed \end{sol}$$
$$\begin{ex} The equation of the plane passing through three points A(6,9,1), B(7,2,3), C(8,5,4) is 13 x-y-10 z-59=0. \end{ex}$$

$$\begin{sol} \mbox{} \begin{itemize} \item We have $\vv{AB} = (7,2,3) - (6,9,1)=(1,-7,2)$ and $\vv{AC} = (8,5,4) - (6,9,1)=(2,-4,3).$ \item Therefore the normal vector to the plane is given by \vv{n}= \vv{AB}\times \vv{AC} = (-13,1,10). \item Equation of the plane is given by $-13 (x-6)+y+10 (z-1)-9=0.$ Expand and simplify, we get 13 x-y-10 z-59=0. \end{itemize} \qed \end{sol}$$
$$\begin{ex} The equation of the plane passing through three points A(6,9,1), B(7,2,4), C(8,3,5) is 5 x-y-4 z-17=0. \end{ex}$$

$$\begin{sol} \mbox{} \begin{itemize} \item We have $\vv{AB} = (7,2,4) - (6,9,1)=(1,-7,3)$ and $\vv{AC} = (8,3,5) - (6,9,1)=(2,-6,4).$ \item Therefore the normal vector to the plane is given by \vv{n}= \vv{AB}\times \vv{AC} = (-10,2,8). \item Equation of the plane is given by $-10 (x-6)+2 (y-9)+8 (z-1)=0.$ Expand and simplify, we get 5 x-y-4 z-17=0. \end{itemize} \qed \end{sol}$$
\end{document}


I am seeing situation like this question