Answer:
The tower deviates
from the vertical.
Explanation:
Having 3 point of our new plane we can construct vectors on the plane by substacting 2 of them:
![v_1=(0,3,2)-(0,0,0)\\v_2=(1,3,0)-(0,0,0)](https://img.qammunity.org/2020/formulas/mathematics/college/iy3jbu7zd1y5wtyqfa3xs4skbiyh6hci89.png)
These vectors are on the plane, so a cross product between them will give us a vector perpendicular to the plane:
![(0,3,2)*(1,3,0)=\left[\begin{array}{ccc}i&j&k\\0&3&2\\1&3&0\end{array}\right] =(-6,2,-3)](https://img.qammunity.org/2020/formulas/mathematics/college/2zq87q0zs33s9szdf23rx0yu80jzqg48l1.png)
Asuming that the aliens used our conventions, the original plane was perpendicular to the z axis, so that a perpendicular vector to that plane was
(0,0,1)
We know that a dot product between 2 vectors |V.W| = |V| |W| cos(α), where α is the angle between them. If we use the vector perpendicular to this plane, and the one perpendicular to the original plane, α will represent the deviation angle of our new plane.
![\|(-6,2,-3)\|=7\\\|(0,0,1)\|=1](https://img.qammunity.org/2020/formulas/mathematics/college/t6eomcffemosg775onmv0wampsie1wom9o.png)
![(-6,2,-3)\odot(0,0,1)= 7 cos(\alpha )\\\\ -3=7 cos(\alpha ) \\ \alpha=arc\ cos (-3)/(7) =115.37 ^\circ\\](https://img.qammunity.org/2020/formulas/mathematics/college/9q5938b0bq7fmhn1gw8igl9smsdddzxswi.png)
Since this angle is greater than 90 degrees it means that the vector we calculated as perpendicular to the plane points towards negative z (this can be seen by the -3 z component)
To fix this we can calculate a new perpendicular vector, or simply compare ir with the vector (0,0,-1). The latter is easier:
![(-6,2,-3)\odot(0,0,-1)= 7 cos(\alpha )\\\\ 3=7 cos(\alpha ) \\ \alpha=arc\ cos (3)/(7) =64.6^\circ =64^\circ36'](https://img.qammunity.org/2020/formulas/mathematics/college/kkhy54gx954jg1oqw5gtob8ca1qace9lvw.png)