Spheres and solid geometry!

Let the coordinates of $S, A, B, C$ be $(0,2,0),(a, b, c), (d, e, f), (0,-2,0)$ respectively. Then the required volume of the tetrahedron is $\dfrac{1}{6}\overrightarrow {CS}\cdot (\overrightarrow {CA}\times \overrightarrow {CB})$ . Since $\overrightarrow {CS}=4\hat j$ , only the $y$ -component of the vector product needs be calculated, which is $cd-af$ . So the volume is $V=\dfrac{2}{3}(cd-af)$ . Again $\overrightarrow {BS}\cdot \overrightarrow {CS}=|\overrightarrow {BS}||\overrightarrow {CS}|\cos 30\degree$ . This implies $4(2-e)=4\sqrt {d^2+(2-e)^2+f^2}\times \dfrac{\sqrt 3}{2}$ , or $(2-e)^2=3(d^2+f^2)$ . Also $d^2+e^2+f^2=4$ . Hence $e^2-e-2=0$ , or $e=-1,2$ . For $e=2$ , we get the point $S$ . So, the coordinates of $B$ are $(d, -1,\sqrt {3-d^2})$ and of $A$ are $(d, -1,-\sqrt {3-d^2})$ . Hence $a=d, c=\sqrt {3-d^2}, f=-\sqrt {3-d^2}$ and $V=\dfrac{2}{3}\times (d\sqrt {3-d^2}+d\sqrt {3-d^2})=\dfrac{4}{3}d\sqrt {3-d^2}\implies V^2=\dfrac{16}{9}d^2(3-d^2)\leq \dfrac{16}{9}(\dfrac{d^2+3-d^2}{2})^2$ . Hence the maximum value of $V^2$ is $\dfrac{16}{9}\times \dfrac{9}{4}=4$ and so the maximum value of $V$ is $\boxed 2$ .