Determinant tactics-101

$\begin{vmatrix} \sin { ({ \alpha }+{ \alpha }_{ 1 }) } \sin { ({ \alpha - }{ \alpha }_{ 1 }) } & \quad \sin { ({ \alpha - }{ \alpha }_{ 1 }) } \cos { ({ \alpha + }{ \alpha }_{ 1 }) } & \quad 2\sin { ({ \alpha }) } \\ \sin { ({ \alpha + }{ \alpha }_{ 3 }) } \sin { ({ \alpha }-{ \alpha }_{ 3 }) } & \quad \sin { ({ \alpha -{ \alpha }_{ 3 } }) } \cos { ({ \alpha + }{ \alpha }_{ 3 }) } & \quad 2\sin { ({ \alpha }) } \\ \sin { ({ \alpha +{ \alpha }_{ 2 } }) } \sin { ({ \alpha - }{ \alpha }_{ 2 }) } & \quad \sin { ({ \alpha -{ \alpha }_{ 2 } }) } \cos { ({ \alpha }+{ \alpha }_{ 2 }) } & \quad 2\sin { ({ \alpha }) } \end{vmatrix}\\ =\frac { 1 }{ 2 } \sin { ({ \alpha }) } \begin{vmatrix} 2\sin { ({ \alpha }+{ \alpha }_{ 1 }) } \sin { ({ \alpha - }{ \alpha }_{ 1 }) } & \quad 2\sin { ({ \alpha - }{ \alpha }_{ 1 }) } \cos { ({ \alpha + }{ \alpha }_{ 1 }) } & \quad 1 \\ 2\sin { ({ \alpha + }{ \alpha }_{ 3 }) } \sin { ({ \alpha }-{ \alpha }_{ 3 }) } & \quad 2\sin { ({ \alpha -{ \alpha }_{ 3 } }) } \cos { ({ \alpha + }{ \alpha }_{ 3 }) } & \quad 1 \\ 2\sin { ({ \alpha +{ \alpha }_{ 2 } }) } \sin { ({ \alpha - }{ \alpha }_{ 2 }) } & \quad 2\sin { ({ \alpha -{ \alpha }_{ 2 } }) } \cos { ({ \alpha }+{ \alpha }_{ 2 }) } & \quad 1 \end{vmatrix}\\ =\frac { \sin { ({ \alpha }) } }{ 2 } \begin{vmatrix} \cos { 2{ \alpha }_{ 1 } } -\cos { 2\alpha } & \sin { 2\alpha } -\sin { 2{ \alpha }_{ 1 } } & 1 \\ \cos { 2{ \alpha }_{ 3 } } -\cos { 2\alpha } & \sin { 2\alpha } -\sin { 2{ \alpha }_{ 3 } } & 1 \\ \cos { 2{ \alpha }_{ 2 } } -\cos { 2\alpha } & \sin { 2\alpha } -\sin { 2{ \alpha }_{ 2 } } & 1 \end{vmatrix}\\ { C }_{ 2 }\longrightarrow { C }_{ 2 }-\sin { 2\alpha } { C }_{ 3 },{ C }_{ 1 }\longrightarrow { C }_{ 1 }+\cos { 2\alpha } { C }_{ 3 }\\ \Rightarrow =\frac { -\sin { ({ \alpha }) } }{ 2 } \begin{vmatrix} \cos { 2{ \alpha }_{ 1 } } & \sin { 2{ \alpha }_{ 1 } } & 1 \\ \cos { 2{ \alpha }_{ 3 } } & \sin { 2{ \alpha }_{ 3 } } & 1 \\ \cos { 2{ \alpha }_{ 2 } } & \sin { 2{ \alpha }_{ 2 } } & 1 \end{vmatrix}$

Now, the determinant taken with the $1/2$ is simply the expression for area of triangle inscribed in unit circle centred at origin. This means that it's maximum value is simply the maximum area of triangle that can be inscribed in a unit circle $=\frac { 3\sqrt { 3 } }{ 4 }$ . (equilateral triangle with side $3^{1/2}$ )

Also, it's easy to see that $max(-\sin { ({ \alpha }) } )=1$

$\Rightarrow X=\frac { 3\sqrt { 3 } }{ 4 }$

$\Rightarrow 48X^{2}=\boxed{81}$