An unusual symmetric inequality of trigonometric functions

Given $\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2$.
I have to prove that $\left| \begin{matrix} \cos\alpha & \cos\beta & \sin\gamma\\\sin\alpha & \cos\beta & \cos\gamma\\\cos\alpha & \sin\beta & \cos\gamma \end{matrix} \right| \leq 2\sqrt2 \sin\alpha \sin\beta \sin\gamma$.
I decided to directly expand the determinant, the left becomes $|2\cos \alpha\cos\beta\cos\gamma-\sin(\alpha+\beta+\gamma)|$. This is quite different from what I encountered before as the equal sign doesn't happen when $\alpha=\beta=\gamma$ like the usual symmetric inequalities.
edit: sorry i forgot the condition that $0\leq\alpha, \beta, \gamma \leq \pi$

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$\alpha$, $\beta$, $\gamma$ , are the angles made by a line with x , y and z axis respectively, then the summation of squares of their sin is 2, and in that case $\alpha$+ $\beta$+ $\gamma$ = 4pi – Tomarinator Mar 12 '12 at 18:53
+1, and I wonder why I'm the only person who's voted in this question's favor when it's got an answer with two up-votes? – Michael Hardy Mar 12 '12 at 19:08
A good answer does not necessary implies it belongs to a equally good question. – Tomarinator Mar 12 '12 at 19:21
@ 5ToM sorry my bad i forgot the condition :) – Geralt of Rivia Mar 12 '12 at 19:52

I assume that these sines are non-negative (otherwise absolute value bars are missing in the RHS). Consider the rows of the matrix as vectors in $\mathbb{R}^3$. Then the lengths of those vectors are $\sqrt{2} \sin(\gamma)$, $\sqrt{2} \sin(\alpha)$ and $\sqrt{2} \sin(\beta)$ respectively. The maximum volume of a parallelepiped with those edge lengths is the product of these lengths (if the rows are orthogonal).