[math]x=-y\ cot\ 29.5^{\circ}+0.09156[/math]
Parameterizing this
[math]r\mapsto {-y\ cot\ 29.5^{\circ}+0.09156,y,0}[/math]
[math]t\mapsto {t\ cos\ 29.5^{\circ}+0.09156,-t\ sin\ 29.5^{\circ},0}[/math]
where the negative sign is applied to the sine function by the even odd relationships of cosine and sine, i.e. ( sin(-t)=-sin(t), cos(-t)=cos(t)) and the fact that the y component is in the 4th quadrant.
[math]
\begin{bmatrix}
x'' \\
y'' \\
z''
\end{bmatrix}=
\begin{bmatrix}
cos\ 6^{\circ} & -sin\ 6^{\circ} & 0 \\
sin\ 6^{\circ} & cos\ 6^{\circ} & 0 \\
0 & 0 & 1
\end{bmatrix} \cdot
\begin{bmatrix}
x' \\
y' \\
z'
\end{bmatrix}[/math]
[math]
\begin{bmatrix}
x'' \\
y'' \\
z''
\end{bmatrix}=
\begin{bmatrix}
cos\ 6^{\circ} & -sin\ 6^{\circ} & 0 \\
sin\ 6^{\circ} & cos\ 6^{\circ} & 0 \\
0 & 0 & 1
\end{bmatrix} \cdot
\begin{bmatrix}
t\ cos\ 29.5^{\circ}+0.09156 \\
-t\ sin\ 29.5^{\circ} \\
0
\end{bmatrix}[/math]
[math]
\begin{bmatrix}
x'' \\
y'' \\
z''
\end{bmatrix}=
\begin{bmatrix}
0.09156\ cos\ 6^{\circ}+t\ cos\ 6^{\circ}cos\ 29.5^{\circ}+t\ sin\ 6^{\circ}sin\ 29.5^{\circ} \\
-t\ cos\ 6^{\circ}sin\ 29.5^{\circ}+0.09156\ sin\ 6^{\circ}+t\ cos\ 29.5^{\circ}sin\ 6^{\circ} \\
0
\end{bmatrix}[/math]
Using the equation for y we can solve for t
[math]y''=0.09156\ sin\ 6^{\circ}-t\ sin\ 23.5^{\circ} \Rightarrow t=\frac{-(y''-0.09156 sin 6^{\circ})}{sin 23.5^{\circ}}[/math]
Substituting this into the expression for x
[math]x''=0.09156\ cos\ 6^{\circ}+t\ cos\ 23.5^{\circ}[/math]
[math]x''=0.09156\ cos\ 6 ^{\circ}+\frac{-(y''-0.09156 sin 6 ^{\circ})}{sin 23.5^{\circ}}(cos 23.5^{\circ})[/math]
[math]x''=0.091058+\frac{y''-.0095706 }{-0.398749} (.917060)[/math]
[math]x''=0.091058+(y''-.0095706 ) (-2.299843)[/math]
[math]x''=-2.299843\ y''+.022011+.091058[/math]
[math]x''=-2.299843\ y''+.113069[/math]