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(D) Given that $\alpha, \beta, \gamma$ are the angles of a triangle,we have $\alpha + \beta + \gamma = 180^{\circ}$.We are given the equations:$(i) \q

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(D) Given that $\alpha, \beta, \gamma$ are the angles of a triangle,we have $\alpha + \beta + \gamma = 180^{\circ}$.We are given the equations:$(i) \quad 2 \sin \alpha + 3 \cos \beta = 3 \sqrt{2}$$(ii) \quad 3 \sin \beta + 2 \cos \alpha = 1$Squaring both equations:$(2 \sin \alpha + 3 \cos \beta)^2 = (3 \sqrt{2})^2 = 18$$4 \sin^2 \alpha + 9 \cos^2 \beta + 12 \sin \alpha \cos \beta = 18$$(3 \sin \beta + 2 \cos \alpha)^2 = 1^2 = 1$$9 \sin^2 \beta + 4 \cos^2 \alpha + 12 \sin \beta \cos \alpha = 1$Adding these two equations:$4(\sin^2 \alpha + \cos^2 \alpha) + 9(\sin^2 \beta + \cos^2 \beta) + 12(\sin \alpha \cos \beta + \cos \alpha \sin \beta) = 18 + 1$$4(1) + 9(1) + 12 \sin(\alpha + \beta) = 19$$13 + 12 \sin(\alpha + \beta) = 19$$12 \sin(\alpha + \beta) = 6$$\sin(\alpha + \beta) = \frac{1}{2}$Since $\alpha + \beta + \gamma = 180^{\circ}$,we have $\alpha + \beta = 180^{\circ} - \gamma$. Thus,$\sin(180^{\circ} - \gamma) = \sin \gamma = \frac{1}{2}$.Therefore,$\gamma = 30^{\circ}$ or $\gamma = 150^{\circ}$.Checking the original equations,if $\gamma = 150^{\circ}$,then $\alpha + \beta = 30^{\circ}$,which is not possible for positive angles $\alpha, \beta$ satisfying the given equations. Thus,$\gamma = 30^{\circ}$.

List-$I$	List-$II$
$(A)$ $\sum \cot A$	$(i)$ $\frac{(a+b+c)^2}{4\Delta}$
$(B)$ $\sum \cot \frac{A}{2}$	$(ii)$ $\frac{a^2+b^2+c^2}{4\Delta}$
$(C)$ If $\tan A : \tan B : \tan C = 1 : 2 : 3$,then $\sin A : \sin B : \sin C =$	$(iii)$ $8 : 6 : 5$
$(D)$ If $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$,then $a : b : c =$	$(iv)$ $12 : 5 : 13$
	$(v)$ $\sqrt{5} : 2\sqrt{2} : 3$
	$(vi)$ $4\Delta$

The angles $\alpha, \beta, \gamma$ of a triangle satisfy the equations $2 \sin \alpha + 3 \cos \beta = 3 \sqrt{2}$ and $3 \sin \beta + 2 \cos \alpha = 1$. Then,angle $\gamma$ equals (in $^{\circ}$)

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