(d)

We have,

$\alpha, \beta, \gamma$ are angle of triangle.

$\therefore \quad \alpha+\beta+\gamma=180^{\circ}$

Given, $\quad 2 \sin \alpha+3 \cos \beta=3 \sqrt{2}$

and $\quad 3 \sin \beta+2 \cos \alpha=1$

On squaring and adding Eqs. $(i)$ and $(ii)$

we get

$4 \sin ^2 \alpha+9 \cos ^2 \beta+12 \sin \alpha \cos \beta$

$+9 \sin ^2 \beta+4 \cos ^2 \alpha$

$+12 \cos \alpha \sin \beta=18+1$

$\Rightarrow 4\left(\sin ^2 \alpha+\cos ^2 \alpha\right)+9\left(\cos ^2 \beta+\sin ^2 \beta\right)$

$+12(\sin \alpha \cos \beta+\cos \alpha \sin \beta)=19$

$\Rightarrow \quad 4+9+12 \sin (\alpha+\beta)=19$

$\Rightarrow \quad 12 \sin (\alpha+\beta)=19-9-4$

$\Rightarrow \sin (\alpha+\beta) =\frac{1}{2}$

$\Rightarrow \alpha+\beta =150^{\circ}$

$\therefore \gamma =30^{\circ}$