$ABCD$ has area $9$. $M$ is in the middle of $AB$ and the edge $BF$ of length $2$ forms an angle of $60º$. Calculate $[CM,CB,BF]$, knowing that $\mathbb{V}^3$ is oriented by a positive basis.
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As the computation of the determinant ammounts to the volume of the paralelogram, I've obtained the height of $GP$ with the trigonometric identity:
$$\frac{2}{\sin 90}=\frac{a}{\sin 60}=\frac{b}{\sin 30}\\a=\sqrt{3}$$
The volume of the cube is given by $height\times width \times depth$ and hence we have:
$$ height\times width \times depth \\ heigth \times 9 = 9 \sqrt{3} $$
But in the book, the answer is $3\sqrt{3}$. What is wrong?