If you want an approximate solution, it's not really simple.

rref(A) would give you quick access to determine what the solutions to the linear system (first 3 equations) are -- either there will be 1 solution (plug it into the sphere equation and see if it's true), or the planes will all meet in a straight line (infinitely many solutions) -- find the line equation and plug it into the sphere equation and solve (that straight line will surely hit the sphere 0, 1, or 2 times), or all 3 of the first equations will correspond to the same plane (no solution or a whole slice of the sphere where the plane cuts through it will solve the system).

You could write code with if statements to handle all of this, but it's more trouble than just doing the intermediate steps by hand.

This post was edited by Amaston on Sep 20 2015 10:22am

nevermind, just found a nice equaiton solver online

http://www.quickmath.com/webMathematica3/quickmath/equations/solve/advanced.jsp#c=solve_advancedsolveequations&v1=-300.21x%2B45y%2B25z+%3D+0%0A45x+-+40.21y+-+65z+%3D+0%0Ax%5E2+%2B+y%5E2+%2B+z%5E2+%3D+1&v2=x%0Ay%0Az

no, this

How r u gonna put x^2 + ysquare + z square = 1 into a matrix along wit the other 3 equation...

how about you start with the 3 equations for the row reduced echelon form, then put the solution into your constraint to see if it works?

It will just come up 0,0,0 no?

its gonna come out to be x=0,y=0,z=0

which obviously doesnt fit the constraint

Damned !

context of problem