1. We need to evaluate

$$\lim_{x\to 0}\frac{e^{ax}-\cos(bx)-\frac{cx e^{-cx}}{2}}{1-\cos(2x)}=17.$$

Since the denominator tends to $0$ as $x\to 0$, the numerator must also vanish up to order $x^2$, and we compare series expansions.

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2. Expand each term around $x=0$.

For $e^{ax}$:

$$e^{ax}=1+ax+\frac{a^2x^2}{2}+O(x^3).$$

For $\cos(bx)$:

$$\cos(bx)=1-\frac{b^2x^2}{2}+O(x^4).$$

For $e^{-cx}$:

$$e^{-cx}=1-cx+\frac{c^2x^2}{2}+O(x^3).$$

So,

$$\frac{cx e^{-cx}}{2}=\frac{cx}{2}\left(1-cx+\frac{c^2x^2}{2}+O(x^3)\right) =\frac{cx}{2}-\frac{c^2x^2}{2}+O(x^3).$$

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3. Substitute into the numerator:

$$e^{ax}-\cos(bx)-\frac{cx e^{-cx}}{2}$$ $$=\left(1+ax+\frac{a^2x^2}{2}+O(x^3)\right)-\left(1-\frac{b^2x^2}{2}+O(x^4)\right)-\left(\frac{cx}{2}-\frac{c^2x^2}{2}+O(x^3)\right).$$

Now simplify:

$$= ax-\frac{cx}{2}+\frac{a^2x^2}{2}+\frac{b^2x^2}{2}+\frac{c^2x^2}{2}+O(x^3).$$

Thus,

$$\text{Numerator}=\left(a-\frac c2\right)x+\frac{a^2+b^2+c^2}{2}x^2+O(x^3).$$

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4. Expand the denominator:

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

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5. For the limit to be finite, the coefficient of $x$ in the numerator must be zero:

$$a-\frac c2=0 \quad \Rightarrow \quad c=2a.$$

Then the numerator becomes

$$\frac{a^2+b^2+(2a)^2}{2}x^2+O(x^3) =\frac{5a^2+b^2}{2}x^2+O(x^3).$$

So the limit is

$$\lim_{x\to 0}\frac{\frac{5a^2+b^2}{2}x^2+O(x^3)}{2x^2+O(x^4)} =\frac{5a^2+b^2}{4}.$$

Given this equals $17$,

$$\frac{5a^2+b^2}{4}=17.$$

Hence,

$$5a^2+b^2=68.$$

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6. Therefore the correct option is

$$\boxed{68}$$

which is Option B.