The Taylor series for $e^x$, $\cos x$, and $e^{-x}$ around $x=0$ are:

$$e^{ax} = 1 + ax + \frac{(ax)^2}{2!} + \ldots,$$

$$\cos(bx) = 1 - \frac{(bx)^2}{2!} + \ldots,$$

$$e^{-cx} = 1 - cx + \frac{(cx)^2}{2!} - \ldots.$$

Substituting these into the limit and simplifying:

$$\lim\limits_{x \rightarrow 0} \frac{(1+ax+\frac{(ax)^2}{2!})-(1-\frac{(bx)^2}{2!})-\frac{cx}{2}(1-(cx)+\frac{(cx)^2}{2!})}{1-\cos 2x}.$$

This simplifies to:

$$\lim\limits _{x \rightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{\left(\frac{1-\cos 2 x}{(2 x)^2}\right) \times 4 x^2}.$$

Let's evaluate this expression:

$$(\frac{1-\cos 2x}{(2x)^2}) \times 4x^2.$$

We know that $1 - \cos 2x$ can be rewritten using the double-angle formula for cosine, which is $\cos 2x = 1 - 2\sin^2x$. So, $1 - \cos 2x = 2\sin^2x$.

Substituting this into our expression we get:

$$(\frac{2\sin^2x}{(2x)^2}) \times 4x^2.$$

This simplifies to:

$${1 \over 2}$$$$(\frac{\sin x}{x})^2 \times 4x^2 = {1 \over 2} \times 4x^2 = 2x^2.$$

So,

$$(\frac{1-\cos 2x}{(2x)^2}) \times 4x^2 = 2x^2.$$

$$ \therefore $$ $$\lim _{x \rightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{2x^2} = 17.$$

Now, for the limit to exist, the coefficient of $x$ in the numerator must be zero, as the limit would be undefined otherwise. This gives us the equation:

$$a-\frac{c}{2} = 0 \Rightarrow c=2a.$$

Then, for the limit to equal $17$, the coefficient of $x^2$ in the numerator must equal $17 \times 2 = 34$. This gives us the equation:

$$\frac{a^2+b^2+c^2}{2} = 34 \Rightarrow a^2+b^2+c^2 = 68.$$

Since $c=2a$, we can substitute $c$ in the equation to get:

$$a^2 + b^2 + (2a)^2 = 68 \Rightarrow 5a^2 + b^2 = 68.$$