Let $\frac { \pi } { 2 } < x < \pi$ be such that $\cot x = \frac { - 5 } { \sqrt { 11 } }$. Then
$$\left( \sin \frac { 11 x } { 2 } \right) ( \sin 6 x - \cos 6 x ) + \left( \cos \frac { 11 x } { 2 } \right) ( \sin 6 x + \cos 6 x )$$
is equal to
(A) $\frac { \sqrt { 11 } - 1 } { 2 \sqrt { 3 } }$
(B) $\frac { \sqrt { 11 } + 1 } { 2 \sqrt { 3 } }$
(C) $\frac { \sqrt { 11 } + 1 } { 3 \sqrt { 2 } }$
(D) $\frac { \sqrt { 11 } - 1 } { 3 \sqrt { 2 } }$