How do you find the range of the scalar triple product of the vectors #<cos alpha, cos beta, 0>, <0, cos beta, cos gamma> and <cos alpha, 0, cos gamma>?#

The triple scalar product of vectors #vec(u), vec(v), vec(w) in RR^3# is given by #vec(u)*(vec(v)xxvec(w))#. A useful fact is that this can also be calculated as the determinant of a #3xx3# matrix with the given vectors as the rows or columns. That is,

#vec(u)*(vec(v)xxvec(w)) = det((u_1, u_2, u_3), (v_1, v_2, v_3), (w_1, w_2, w_3))#

We will use that to calculate the given scalar triple product.

Let
#vec(u) = < cos(alpha), cos(beta), 0 >#
#vec(v) = < 0, cos(beta), cos(gamma) >#
#vec(w) = < cos(alpha), 0, cos(gamma) >#

Then we have:

#vec(u)*(vec(v)xxvec(w)) = det((u_1, u_2, u_3), (v_1, v_2, v_3), (w_1, w_2, w_3))#

#=det((cos(alpha),cos(beta),0),(0,cos(beta),cos(gamma)),(cos(alpha),0,cos(gamma)))#

#=cos(alpha)(cos(beta)cos(gamma)-0cos(gamma))#

#-cos(beta)(0cos(gamma)-cos(alpha)cos(gamma))#

#+0(0*0-cos(alpha)cos(beta))#

#=2cos(alpha)cos(beta)cos(gamma)#

Thus, it suffices to find the range of #2cos(alpha)cos(beta)cos(gamma)#.

With no restrictions on #alpha, beta, gamma# beyond being in #RR#, we have #cos(alpha), cos(beta), cos(gamma) in [-1, 1]#.
Thus their product will also have the range #cos(alpha)cos(beta)cos(gamma) in [-1, 1]#.
Multiplying by #2# gives our desired range:
#2cos(alpha)cos(beta)cos(gamma) in [-2, 2]#

Thus the range of the scalar triple product of #< cos(alpha), cos(beta), 0 >, < 0, cos(beta), cos(gamma) >,# and #< cos(alpha), 0, cos(gamma) >#, is #[-2, 2]#