Tilted cuboid tank - 2

Freezing the water surface with respect to the cuboid, and un-tilting the cuboid (i.e. returning it to the upright orientation it started with), we note that the water surface must pass though the top edge point $\mathbf{r}_1 = (\frac{5}{2}, -\frac{7}{2}, 9 )$ . From symmetry, it also must pass through the point $\mathbf{r}_2 = (0, 0, 6)$ . The unit normal to the water surface before un-tilting is $(0, 0, 1)$ . So after un-tilting, it will point in the direction $\mathbf{n} = ( - \sin 60^{\circ} \sin \theta, \cos 60^{\circ} \sin \theta , \cos \theta )$ . Now we can write,

$\mathbf{n} \cdot (\mathbf{r}_1 - \mathbf{r}_2 ) = 0$

Substituting $\mathbf{n} , \mathbf{r}_1 , \mathbf{r}_2$ ,

$( - \sin 60^{\circ} \sin \theta, \cos 60^{\circ} \sin \theta , \cos \theta ) \cdot (\frac{5}{2}, -\frac{7}{2}, 3 ) = 0$

Multiplying by $4$ throughout, and simplifying,

$\sin \theta ( -5 \sqrt{3} - 7 ) + 12 \cos \theta = 0$

So that,

$\tan \theta = \dfrac{12}{5 \sqrt{3} + 7 }$

From which, it follows that,

$\theta = 37.4619024051543...$

And this makes the answer $\boxed{374619}$