Given,

$$F(x) = \left| {\matrix{ {a + \sin {x \over 2}} & { - b\cos x} & 0 \cr { - b\cos x} & 0 & {a + \sin {x \over 2}} \cr 0 & {a + \sin {x \over 2}} & { - b\cos x} \cr } } \right|$$

$$ = \left( {a + \sin {x \over 2}} \right)\left( { - {{\left( {a + \sin {x \over 2}} \right)}^2}} \right) + b\cos x \times {b^2}{\cos ^2}x$$

$$ = - {\left( {a + \sin {x \over 2}} \right)^3} - {b^3}{\cos ^3}x$$

Now,

$$\mathop {\lim }\limits_{x \to 0} {{F(x)} \over {{x^3}}}$$

$$ = \mathop {\lim }\limits_{x \to 0} {{ - {{\left( {a + \sin {x \over 2}} \right)}^3} - {b^3}{{\cos }^3}x} \over {{x^3}}}$$

Given limit exists, it only possible when a = 0 and b = 0.

$$ = \mathop {\lim }\limits_{x \to 0} {{ - {{\left( {\sin {x \over 2}} \right)}^3}} \over {{x^3}}}$$

$$ = \mathop {\lim }\limits_{x \to 0} - {\left( {{1 \over 2} \times \left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)} \right)^3}$$

$$ = \mathop {\lim }\limits_{x \to 0} - {\left( {{1 \over 2}} \right)^3} \times {\left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)^3}$$

$$ = - {1 \over 8} \times 1 = L$$

$$\therefore$$ $$ - 112L = - 112 \times - {1 \over 8} = 14$$