If $(2,3,5)$ is one end of a diameter of the sphere $x^2+y^2+z^2-6x-12y-2z+20=0$, then the coordinates of the other end of the diameter are:

The general equation of a sphere is $x^2+y^2+z^2+2ux+2vy+2wz+d=0$.The centre of the sphere is $(-u,-v,-w)$.For the given sphere $x^2+y^2+z^2-6x-12y-2z+20=0$:$2u=-6⟹u=-3$$2v=-12⟹v=-6$$2w=-2⟹w=-1$ So the centre $C$ is $(3,6,1)$.Let the other end of the diameter be $(x_1,y_1,z_1)$. The centre is the midpoint of the diameter.One end is $(2,3,5)$.
$$\frac{x_1+2}{2}=3⟹x_1=6-2=4$$
$$\frac{y_1+3}{2}=6⟹y_1=12-3=9$$
$$\frac{z_1+5}{2}=1⟹z_1=2-5=-3$$
The coordinates of the other end are $(4,9,-3)$.