You're looking for the extreme values of [tex]T(x,y,z)=3x+6y-6z+1[/tex] subject to [tex]x^2+y^2+z^2=9[/tex]. The Lagrangian is
[tex]L(x,y,z,\lambda)=3x+6y-6z+1+\lambda(x^2+y^2+z^2-9)[/tex]
with critical wherever the partial derivatives vanish:
[tex]L_x=3+2\lambda x=0\implies x=-\dfrac3{2\lambda}[/tex]
[tex]L_y=6+2\lambda y=0\implies y=-\dfrac3\lambda[/tex]
[tex]L_z=-6+2\lambda z=0\implies z=\dfrac3\lambda[/tex]
[tex]L_\lambda=x^2+y^2+z^2-9=0[/tex]
Substituting the first three solutions into the last equation gives
[tex]\dfrac9{4\lambda^2}+\dfrac9{\lambda^2}+\dfrac9{\lambda^2}=9[/tex]
[tex]\implies\lambda=\pm\dfrac32[/tex]
[tex]\implies x=1,y=2,z=-2\text{ or }x=-1,y=-2,z=2[/tex]
At these points, we have
[tex]T(1,2,-2)=28[/tex]
[tex]T(-1,-2,2)=-26[/tex]
so the highest temperature the bee can experience is 28º F at the point (1, 2, -2), and the lowest is -26º F at the point (-1, -2, 2).