I think we need a lot more info about this to see how the normality comes into it. With any problem like this it also is unlikely that we are going to be able to solve it without exploring some specific examples. To begin with as well, I tried simplifying the number of dimensions.
First of all, yes, do we mean a tetrahedron or a triangle?
Second I assume, as I mentioned once before, that the sphere is of radius R, centred on the origin
Thirdly, about the normality. What role does the normality play in defining the points on the sphere? A point in 3 dimensional space has in a Cartesian world, 3 parameters that must be specified, not one. So if the coordinates are to be drawn from a normal distribution, then we need 9 (triangle) or 12 (tetrahedron) draws from the (presumably same) distribution. Unless as Ian suggests, we allow ourselves to draw the first point. I assume this distribution is standard normal? In which case what is the value of R, since the distribution is infinitely wide?
Leaving aside the normality for the moment, I wondered about reducing the number of dimensions to get a feel for how this sort of thing operates. Let us assume we have a triangle inside a circle, centre O. What criteria must be satisfied for the triangle to include O? Well, this is simple, because it only requires that the triangle must have all 3 angles less than or equal to 90 degrees. We seem to be in the realm of Euler's circle formulae.
If however we start with a point, and then randomly select another point, so that we have a chord, then conditional on the length of the chord C, we can always achieve the objective of having O included, since the perpendicular bisector of the chord passes through the centre. Let P1 be the bisection point and P2 the point where it cuts the circle after passing through O. Now let us consider any bisector of the chord. It seems as though we can reduce the problem to a single dimension conditional on the length of C, by considering the probability that a randomly chosen point P1 on C will project to a suitable point P2 on the circle?
Unfortunately given the low cost effectiveness of this thinking in terms of what I'm supposed to be doing, I had better go get some tea and come back later