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(A) Let $x = \sin \alpha$, $y = \cos \beta$, and $z = 	an \gamma$. The system becomes:$2x - y + 3z = 3 \quad \dots (i)$$4x + 2y - 2z = 2 \quad \dots

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$2 \alpha - \beta - \gamma = 0$

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(A) Let $x = \sin \alpha$, $y = \cos \beta$, and $z = 	an \gamma$. The system becomes:$2x - y + 3z = 3 \quad \dots (i)$$4x + 2y - 2z = 2 \quad \dots (ii)$$6x - 3y + z = 9 \quad \dots (iii)$Using matrix form $AX = B$, where $A = \begin{bmatrix} 2 & -1 & 3 \ 4 & 2 & -2 \ 6 & -3 & 1 \end{bmatrix}$, $X = \begin{bmatrix} x \ y \ z \end{bmatrix}$, and $B = \begin{bmatrix} 3 \ 2 \ 9 \end{bmatrix}$.The determinant $|A| = 2(2 - 6) - (-1)(4 + 12) + 3(-12 - 12) = 2(-4) + 1(16) + 3(-24) = -8 + 16 - 72 = -64$.Solving for $X = A^{-1}B$, we find $x = 1$, $y = -1$, $z = 0$.Thus, $\sin \alpha = 1 \implies \alpha = \pi/2$.$\cos \beta = -1 \implies \beta = \pi$.$	an \gamma = 0 \implies \gamma = 0$.Checking the options: $2\alpha - \beta - \gamma = 2(\pi/2) - \pi - 0 = \pi - \pi = 0$. Hence, option $A$ is correct.

For $\alpha, \beta \in [0, 2\pi]$ and $\gamma \in [0, \pi)$, consider the system of equations:
$2 \sin \alpha - \cos \beta + 3 \tan \gamma = 3$
$4 \sin \alpha + 2 \cos \beta - 2 \tan \gamma = 2$
$6 \sin \alpha - 3 \cos \beta + \tan \gamma = 9$
Then, which one of the following is true?

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For $\alpha, \beta \in [0, 2\pi]$ and $\gamma \in [0, \pi)$, consider the system of equations:$2 \sin \alpha - \cos \beta + 3 \tan \gamma = 3$$4 \sin \alpha + 2 \cos \beta - 2 \tan \gamma = 2$$6 \sin \alpha - 3 \cos \beta + \tan \gamma = 9$Then, which one of the following is true?