Answer
The equation of the plane containing the satellite's orbit:
$20x - 29y + 9z = 0$
Work Step by Step
Evaluate the vector ${\bf{J}}$ at initial condition:
${\bf{J}} = {\bf{r}}\left( t \right) \times {\bf{r}}'\left( t \right)$
${\bf{J}} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{29000}&{20000}&0\\
1&1&1
\end{array}} \right|$
${\bf{J}} = 20000{\bf{i}} - 29000{\bf{j}} + 9000{\bf{k}}$
Since the plane requested is orthogonal to ${\bf{J}}$, the normal vector to the plane is parallel to ${\bf{J}}$.
By Eq. (4) of Theorem 1 (Section 13.5), the equation of the plane is
$20000\left( {x - {x_0}} \right) - 29000\left( {y - {y_0}} \right) + 9000\left( {z - {z_0}} \right) = 0$
$20\left( {x - {x_0}} \right) - 29\left( {y - {y_0}} \right) + 9\left( {z - {z_0}} \right) = 0$
We choose the initial position as the point on the plane: $\left( {{x_0},{y_0},{z_0}} \right) = \left( {29000,20000,0} \right)$ and substitute it in the equation above to give
$20\left( {x - 29000} \right) - 29\left( {y - 20000} \right) + 9\left( {z - 0} \right) = 0$
$20x - 29y + 9z = 0$
Thus, this is the equation of the plane containing the satellite's orbit.
