1. Write the system in matrix form

The system is

We solve it for integer triples $(x,y,z)$.

2. Solve the linear system

From the first equation,

Substitute into the second equation: $-2x+4y+z=0.$ Using $x=2y-5z$, $-2(2y-5z)+4y+z=0$ $-4y+10z+4y+z=0$ $11z=0 \implies z=0.$

Then from $x=2y-5z$, $x=2y.$

Now check the third equation: $-7x+14y+9z=-7(2y)+14y+0=0,$ so it is automatically satisfied.

Hence all solutions are of the form $(x,y,z)=(2y,y,0).$ Let $y=t\in \mathbb Z$. Then $(x,y,z)=(2t,t,0).$

3. Apply the condition on $x^2+y^2+z^2$

We need $15\le x^2+y^2+z^2\le 150.$

For $(x,y,z)=(2t,t,0)$, $x^2+y^2+z^2=(2t)^2+t^2+0=4t^2+t^2=5t^2.$

So the condition becomes $15\le 5t^2\le 150.$ Divide by $5$: $3\le t^2\le 30.$

Now $t$ is an integer, so possible values of $t^2$ are $4,9,16,25.$ Thus $t=\pm 2,\ \pm 3,\ \pm 4,\ \pm 5.$

That gives $8$ integer solutions.

4. Conclusion

Therefore, the number of elements in the set $S$ is $\boxed{8}.$