= Solution
Use axes adapted to the two orthogonal rotations in the diagram:
$$
\mathbf n_A=\hat{\mathbf z},\qquad \mathbf n_B=\frac{\hat{\mathbf x}+\hat{\mathbf z}}{\sqrt2},\qquad \mathbf n_C=\frac{\hat{\mathbf y}+\hat{\mathbf z}}{\sqrt2}.
$$
For a linearly oscillating <Hertzian dipole> with real unit axis $\mathbf d$ and total time-averaged radiated <power> $P$, the <angular distribution of electric dipole radiation> is
$$
I(\mathbf n)=\frac{3P}{8\pi R^2}\bigl[1-(\mathbf d\cdot\mathbf n)^2\bigr].
$$
The normalization follows from $\int\sin^2\vartheta\,d\Omega=8\pi/3$. Equal readings therefore imply
$$
(d_x+d_z)^2=2d_z^2,\qquad (d_y+d_z)^2=2d_z^2.
$$
Here $d_z\neq0$, since otherwise all components would vanish. Reversing the axis is immaterial, so choose $d_z>0$. The complete family is
$$
\boxed{\mathbf d\parallel(s_B\sqrt2-1,\ s_C\sqrt2-1,\ 1),\qquad s_B,s_C\in\{+1,-1\}.}
$$
Define $a=\sqrt2-1$, $b=-\sqrt2-1$, and $D=1+(s_B\sqrt2-1)^2+(s_C\sqrt2-1)^2$. Then $d_z^2=1/D$, and the absolute <intensity> fixes
$$
P=\frac{8\pi R^2I}{3}\frac{D}{D-1}.
$$
For the measured distance and <intensity>, the allowed axes and powers are:
* $\mathbf d\parallel(a,a,1)$: $D=7-4\sqrt2$, $\boxed{P=32.79\,\mathrm{kW}}$.
* $\mathbf d\parallel(a,b,1)$ or $(b,a,1)$: $D=7$, $\boxed{P=9.774\,\mathrm{kW}}$ for either axis.
* $\mathbf d\parallel(b,b,1)$: $D=7+4\sqrt2$, $\boxed{P=9.096\,\mathrm{kW}}$.
\b[The <intensity> data do not select a unique orientation or total <power>.] This is the <dipole-axis ambiguity from three intensity measurements>: the readings constrain squared projections and lose their signs. For example, the first branch makes an angle $30.36^\circ$ with $A$, but the other branches are equally consistent with the three measurements. A further suitable observation direction or polarization measurement is needed to remove the ambiguity. The calculation uses radiation-zone <intensities>, as appropriate to the <Hertzian dipole> model.
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