Let $u=a+\sin\frac{x}{2},\qquad v=-b\cos x.$ Then $F(x)=\begin{vmatrix}u&v&0\\ v&0&u\\ 0&u&v\end{vmatrix}.$

We are given that $\lim_{x\to 0}\frac{F(x)}{x^3}$ exists and equals $L$.

We must find $-112L$.

1. Compute the determinant

Expand along the first row: \begin{align*} F(x)&=u\begin{vmatrix}0&u\ u&v\end{vmatrix}-v\begin{vmatrix}v&u\ 0&v\end{vmatrix}+0 \ &=u(0\cdot v-u\cdot u)-v(v\cdot v-0)\ &=-u^3-v^3. \end{align*} Since $v=-b\cos x$, $v^3=(-b\cos x)^3=-b^3\cos^3 x.$ Hence

So

2. Use the condition that $\dfrac{F(x)}{x^3}$ has a finite limit

For the limit to exist finitely, we need $F(0)=0,$ and also the coefficients of $x$ and $x^2$ in the expansion of $F(x)$ must vanish.

At $x=0$, $\sin\frac x2=0,\qquad \cos x=1,$ so

Let $a=b=c$. Then

Now expand near $x=0$.

3. Series expansions

We use: $\sin\frac x2=\frac x2-\frac{x^3}{48}+O(x^5),$ $\cos x=1-\frac{x^2}{2}+O(x^4),$ so $\cos^3 x=1-\frac{3x^2}{2}+O(x^4).$

Thus $c^3\cos^3 x=c^3-\frac{3c^3}{2}x^2+O(x^4).$

Now $(c+\sin\tfrac x2)^3=c^3+3c^2\sin\frac x2+3c\sin^2\frac x2+\sin^3\frac x2.$ Using $\sin\frac x2=\frac x2-\frac{x^3}{48}+O(x^5),$ $\sin^2\frac x2=\frac{x^2}{4}+O(x^4),$ $\sin^3\frac x2=\frac{x^3}{8}+O(x^5),$ we get \begin{align*} (c+\sin\tfrac x2)^3 &=c^3+3c^2\left(\frac x2-\frac{x^3}{48}\right)+3c\left(\frac{x^2}{4}\right)+\frac{x^3}{8}+O(x^4)\ &=c^3+\frac{3c^2}{2}x+\frac{3c}{4}x^2+\left(-\frac{3c^2}{48}+\frac18\right)x^3+O(x^4)\ &=c^3+\frac{3c^2}{2}x+\frac{3c}{4}x^2+\left(-\frac{c^2}{16}+\frac18\right)x^3+O(x^4). \end{align*}

Therefore \begin{align*} F(x)&=\left(c^3-\frac{3c^3}{2}x^2+O(x^4)\right) -\left(c^3+\frac{3c^2}{2}x+\frac{3c}{4}x^2+\left(-\frac{c^2}{16}+\frac18\right)x^3+O(x^4)\right)\ &=-\frac{3c^2}{2}x+\left(-\frac{3c^3}{2}-\frac{3c}{4}\right)x^2+\left(\frac{c^2}{16}-\frac18\right)x^3+O(x^4). \end{align*}

For $F(x)/x^3$ to have a finite limit, coefficients of $x$ and $x^2$ must be zero.

So, $-\frac{3c^2}{2}=0\implies c=0.$ This also makes the $x^2$ coefficient zero.

Hence $a=b=0.$

4. Compute $L$

If $a=b=0$, then $F(x)=-(\sin\tfrac x2)^3.$ Therefore \begin{align*} L&=\lim_{x\to 0}\frac{-(\sin\tfrac x2)^3}{x^3} = -\left(\lim_{x\to 0}\frac{\sin(x/2)}{x}\right)^3. \end{align*} Now $\frac{\sin(x/2)}{x}=\frac12\cdot \frac{\sin(x/2)}{x/2}\to \frac12.$ So $L=-\left(\frac12\right)^3=-\frac18.$

Hence $-112L=-112\left(-\frac18\right)=14.$

5. Final answer

$\boxed{14}$