1. We need

$$\lim_{x\to 0}\frac{x^2\sin(\alpha x)+(\gamma-1)e^{x^2}}{\sin 2x-\beta x}=3.$$

For this limit to be finite and equal to $3$, both numerator and denominator must approach $0$ as $x\to 0$.

2. First examine the denominator:

$$\sin 2x-\beta x \to 0-0=0$$

as $x\to 0$, but more precisely,

$$\sin 2x = 2x-\frac{(2x)^3}{3!}+\cdots = 2x-\frac{4}{3}x^3+\cdots$$

Hence

$$\sin 2x-\beta x = (2-\beta)x-\frac{4}{3}x^3+\cdots$$

3. Now examine the numerator:

$$x^2\sin(\alpha x)+ (\gamma-1)e^{x^2}.$$

As $x\to 0$,

$$x^2\sin(\alpha x)\to 0,$$

and

$$e^{x^2}\to 1.$$

So numerator tends to

$$\gamma-1.$$

For the limit to be finite while denominator tends to $0$, we must have

$$\gamma-1=0 \implies \gamma=1.$$

4. With $\gamma=1$, numerator becomes

$$x^2\sin(\alpha x).$$

Using expansion,

$$\sin(\alpha x)=\alpha x-\frac{\alpha^3x^3}{6}+\cdots$$

Therefore

$$x^2\sin(\alpha x)=\alpha x^3+\cdots$$

So numerator is of order $x^3$.

5. For the quotient to approach a finite nonzero number $3$, denominator must also be of order $x^3$. Hence the coefficient of $x$ in denominator must vanish:

$$2-\beta=0 \implies \beta=2.$$

Then

$$\sin 2x-\beta x = \sin 2x-2x = -\frac{4}{3}x^3+\cdots$$

6. Now compute the limit:

$$\lim_{x\to 0}\frac{x^2\sin(\alpha x)}{\sin 2x-2x} =\lim_{x\to 0}\frac{\alpha x^3+\cdots}{-\frac{4}{3}x^3+\cdots} =\frac{\alpha}{-4/3} = -\frac{3\alpha}{4}.$$

This is given to be $3$, so

$$-\frac{3\alpha}{4}=3 \implies \alpha=-4.$$

7. Therefore,

$$\beta+\gamma-\alpha=2+1-(-4)=7.$$

8. So the correct option is

$$\boxed{7}.$$