<s>[INST] new plane equation after transformation of coordinates I have a plane equation $ax + by + cz + d = 0$ w.r.t to a particular coordinate frame. 
this coordinate frame w.r.t to the world coordinate frame is 
$$\begin{vmatrix}
  r_1 & r_2 & r_3 & x_1 \\
  r_4 & r_5 & r_6 & y_1 \\
  r_7 & r_8 & r_9 & z_1 
\end{vmatrix}$$
now, what is the equation of plane in the world coordinate frame?
 [/INST] The equation of the plane can be created from the dot product of two homogeneous vectors
$$ ax+by+cz+d = 0$$
$$ \begin{pmatrix} x\\y\\z\\1 \end{pmatrix}^\intercal \begin{pmatrix} a\\b\\c\\d \end{pmatrix} =0 $$
The world coordinates of the local $(x,y,z,1)$ are
$$ \begin{pmatrix} u \\ v \\ w \\ 1 \end{pmatrix}  = 
\begin{vmatrix}
  r_1 & r_2 & r_3 & x_1 \\
  r_4 & r_5 & r_6 & y_1 \\
  r_7 & r_8 & r_9 & z_1 \\
   0 &   0  &  0  &  1 
\end{vmatrix} \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} $$
Taking the inverse you find that the coefficients $(A,B,C,D)$ of the plane $Au+Bv+Cw+D=0$ are
$$ \begin{pmatrix} A \\ B \\ C\\ D \end{pmatrix} = \begin{vmatrix}
  r_1 & r_2 & r_3 & x_1 \\
  r_4 & r_5 & r_6 & y_1 \\
  r_7 & r_8 & r_9 & z_1 \\
   0 &   0  &  0  &  1 
\end{vmatrix}^{-\intercal} \begin{pmatrix} a\\b\\c\\d \end{pmatrix} $$
</s>