Let $f$ be a differentiable function and the equation of the normal to the graph of $y = f(x)$ at $x = 3$ is $3y = x + 18$. If $L = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( {3 + {{\left( {4{{\tan }^{ - 1}}x - \pi } \right)}^2}} \right) - f\left( {3 + {{\left( {f\left( 3 \right) - x - 6} \right)}^2}} \right)}}{{{{\sin }^2}\left( {x - 1} \right)}}$,then:

• A
  $L = f(-f'(3))$
• B
  $L = 0$
• C
  $L - f'(3) = -6$
• D
  $f(f'(3) + 6) = 9$