Explanation
$$L=\lim _{ x\rightarrow { \pi }/{ 6 } }{ \dfrac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } -1 } } \\ L=\dfrac { 2\sin ^{ 2 }{ \left( \dfrac { \pi }{ 6 } \right) } +\sin { \left( \dfrac { \pi }{ 6 } \right) } -1 }{ 2\sin ^{ 2 }{ \left( \dfrac { \pi }{ 6 } \right) } -3\sin { \left( \dfrac { \pi }{ 6 } \right) } -1 } \\ L=\dfrac { 2.\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 2 } -1 }{ 2.\dfrac { 1 }{ 4 } -3.\dfrac { 1 }{ 2 } -1 } =\dfrac { 0 }{ -2 }= 0 $$
So option $$D$$ is correct
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