Different Ways of Integrating $3\sin x\cos x$

You're forgetting that an indefinite integral must include a constant of integration; for any chosen constant $C$, we have that $$\frac{d}{dx}\left(-\frac{3}{4}\cos(2x)+C\right)=3\sin(x)\cos(x),$$ and that is precisely the relationship captured by the statement that

$$\int 3\sin(x)\cos(x)\,dx=-\frac{3}{4}\cos(2x)+C.$$

All three answers are correct provided you add a constant to each one of those.

Because from the very definition of integration, it is the area under the curve, so it requires bounds to give a unique value. You can't evaluate the value of an indefinite integral without including constant.

And I am sure that in the examination, your son won't be asked to evaluate the value of an integral without providing limits of integration or providing its value at some other point.

For instance, in question it may be mentioned that evaluate the value of expression at x=π/6 , given its value at x=0 is 1. So in this case, all three answers will give the correct value i.e. 11/8