Given the attached figure, we are interested in finding the volume of this red pyramid ($ABCV$) with the spherical base. We assume that the Cartesian coordinates of the points $A, B, C, V$ are known and consequently,
$$A = (x_A, y_A, z_A), B= (x_B, y_B, z_B), C = (x_C, y_C, z_C), V = (x_V, y_V, z_V)$$
The radius $r$ of the sphere is also assumed to be known.
Edit: Since one can easily obtain the plane equation of $ABC$, the desired volume of the red pyramidoid is obtained as: volume of large pyramid $ABCO$ + volume of small pyramid $ABCV$ - volume of the "spherical triangle - like" $ABCO$.
So the new question is: how can we compute the volume of this "spherical pyramidoid" $ABCO$?
Any help would be useful.
You need not worry about the pyramidoid outside the sphere.
Area of spherical triangle $ A_{sph\;triangle} $ is found by spherical excess times r^2.
$$( \alpha + \beta + \gamma - \pi) r^2$$
These three are the dihedral angles at the three vertices.The three equitorial planes have three intersections with the sphere as great circles.
The volume of the convex "pyramidoid" is
$$ V= A_{sph\;triangle} \cdot \dfrac{r}{3}$$ It can be found by integrating thin differential shell elements.
Check for pyramid with very small solid angle at vertex $\to 0$
$$ V=(\pi/3+\pi/3+\pi/3-\pi) r^2 \cdot r/3 \to 0 $$
Check for hemisphere
$$ V=(\pi+\pi+\pi-\pi) r^2 \cdot r/3 = \frac23 \pi r^3 $$
Check for full sphere
$$ V=(\pi+\pi+\pi-\pi) r^2 \cdot r/3 = \frac43 \pi r^3 $$