When the line meets

Since at $x=\frac{π}{2}, x\sin x=\frac{π}{2}>\frac{3}{2}$ and in the interval $(0,π), x\sin x$ is always positive, with a value $0$ at $x=π$ , the equation $x\sin x=\frac{3}{2}$ has two solutions

Solution to this problem :

Let the position coordinates of $A, E, F, M$ be $(0,0,0),(4,\sqrt 2, 0),(2,0,0),(p, q, r)$ respectively.

Then $|\overline {AM}|^2=|\overline {AD}|^2=16\implies p^2+q^2+r^2=4\sqrt 2q+8$

$|\overline {EM}|^2=|\overline {ED}|^2=2\implies p^2+q^2+r^2=8p+2\sqrt 2q-16$ .

$\vec {ME}\perp \vec {MF}\implies p^2+q^2+r^2=6p+\sqrt 2 q-8$ .

Solving these we get $p=\frac{10}{3}, q=\frac{2\sqrt 2}{3}, r=\frac{2}{\sqrt 3}$ .

Unit vector along the normal to the $AEM$ plane is $\hat n=\frac{\vec {AM}\times \vec {AE}}{|\vec {AM}\times \vec {AE}|}$

$=\frac{1}{4\sqrt 2}\left (2\sqrt {\frac{2}{3}}\hat i+\frac{8}{\sqrt 3}\hat j+2\sqrt 2\hat k\right )$

If the required angle be $α$ , then

$\cos α=\hat n. \hat k=\frac{2\sqrt 2}{4\sqrt 2}=\frac{1}{2}\implies α=\boxed {60\degree}$ .