The problem
In the regular quadrilateral prism, $ABCDA'B'C'D'$ the edge of the base is equal to $4 \sqrt{6}$, and the volume is $1152$. Determine the position of the point M on the edge CC', so that the planes $(BA'D)$ and $(MBD)$ are perpendicular.
The idea
drawing
I let O be the center of square ABCD.
If those planes are perpendicular, then the angle they form is 90.
Using the theorem of the 3 perpendicular we get that $MO\perp DB, A'0\perp BD$ and the common side of both planes $(BA'D)$ and $(MBD)$, which make the angle between them be $\angle A'OM=90$
We can calculate using the volume and the edge of the base that the lateral edge is $12$, so AA'=12 and we can also calculate $AO=4\sqrt3$, from here we get that $\angle A0A'=60=> \angle COM=30, => MC=4 $ so we determined the pos of M
The thing is I don't know if this is the only solution... I don't know if M can be outside of CC' Thank you!