Magnitude of F about line CD (Mechanics, 3D vector system) I am having trouble with this question:

I will write out the question below to make it easier for other people googling the question to find it:
2/139 If the magnitude of the moment of F about line CD is 50 Nm, determine the magnitude of F.
I know that $\vec M=\vec r \times \vec F$
What I have tried:
Calculating $\vec M$ as following:
$\vec {CD} = (0.4, 0, -0.2)$ please note how I set up coordinate system (x and y is in bottom and z upward according to right-hand rule)
$|\vec{CD}|$=$\frac{1}{\sqrt{0.4^2+0.2^2}}=2.2$
$\vec M = 50 \times 2.2 (0.4, 0, -0.2)$=$(44, 0, -22)Nm$
I now set $\vec F=(f_1, f_2, f_3)$
then try to find some vector $\vec r$. I have tried $\vec CA$ and $\vec DA$ but no matter what vector I choose I end up with something like $0.2f_1=44$ $0.4f_1=0$ after cross product, which is an impossible system of equation to solve. What am I doing wrong? If $\vec r$ is wrong, please explain why. I am having a hard time wrapping my head around how $\vec F$ can even make a rotation around $\vec CD$ in the first place but I guess this is more of a theoretical question... 
 A: Your issue is that you know the direction of the vector $\boldsymbol{F}$ and therefore $f_1$, $f_2$ and $f_3$ are not independent. Use just one unknown $F$ along the correct direction and the problem becomes solvable.
Consider the line about which the moment is measured with the direction
$$ \boldsymbol{e} = \frac{ \boldsymbol{r}_D - \boldsymbol{r}_C }{ \| \boldsymbol{r}_D - \boldsymbol{r}_C \| } = \frac{ \pmatrix{0.4 \\ -0.2 \\ 0}}{ \tfrac{1}{\sqrt{5}}} = \pmatrix{2/\sqrt{5} \\ -1/\sqrt{5} \\ 0} $$
and the force vector
$$ \boldsymbol{F} =  \frac{ \boldsymbol{r}_B - \boldsymbol{r}_A }{ \| \boldsymbol{r}_B - \boldsymbol{r}_A \| }\,F = \pmatrix{2/\sqrt{6}\\ 1/\sqrt{6} \\ 
 -1/\sqrt{6}} F $$
The moment vector about point C is
$$ \boldsymbol{M} = (\boldsymbol{r}_A - \boldsymbol{r}_C) \times \boldsymbol{F} = \pmatrix{0 \\ -1/5 \\ -1/5} \times \pmatrix{2/\sqrt{6}\\ 1/\sqrt{6} \\ 
 -1/\sqrt{6}} F = \pmatrix{\sqrt{6}/15 \\ -\sqrt{6}/15 \\ \sqrt{6}/15 } F$$
The projection of the moment along the line is supposed to be 50
$$ 50 = \boldsymbol{e} \cdot \boldsymbol{M} =\pmatrix{2/\sqrt{5} \\ -1/\sqrt{5} \\ 0}  \cdot \pmatrix{\sqrt{6}/15 \\ -\sqrt{6}/15 \\ \sqrt{6}/15 }  F = \frac{\sqrt{30}}{25} F $$
$$\Rightarrow\; F = \frac{125 \sqrt{30}}{3} = 228.22 $$
I used the following coordinate system.

A quick shortcut to this calculation is the vector triple product
$$ M = \boldsymbol{e} \cdot ( \boldsymbol{r}_A-\boldsymbol{r}_C) \times \boldsymbol{F} $$
$$ 50 = \pmatrix{2/\sqrt{5} \\ -1/\sqrt{5} \\ 0} \cdot \left[ \pmatrix{0 \\ -1/5 \\ -1/5} \times \pmatrix{2/\sqrt{6}\\ 1/\sqrt{6} \\ 
 -1/\sqrt{6}} \right] F $$ $$ \boxed{ 50 = \frac{\sqrt{30}}{25} F }$$
