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 written in the completed-square form to read its centre and radius.

Start with the given sphere
$$x^{2}+y^{2}+z^{2}-6x-12y-2z+20=0$$

Complete the squares term-wise:

$$x^{2}-6x \;=\;(x-3)^{2}-9$$
$$y^{2}-12y \;=\;(y-6)^{2}-36$$
$$z^{2}-2z \;=\;(z-1)^{2}-1$$

Insert these back into the equation:

$$(x-3)^{2}-9 + (y-6)^{2}-36 + (z-1)^{2}-1 + 20 = 0$$

Simplify the constants:

$$(x-3)^{2} + (y-6)^{2} + (z-1)^{2} = 26$$

Hence the centre of the sphere is $$C(3,\,6,\,1)$$ and the radius is $$\sqrt{26}\,.$

Let $$P(2,\,3,\,5)$$ be one end of a diameter and $$Q(x,\,y,\,z)$$ be the other end. For any diameter, the centre is the midpoint of its endpoints, so

$$C = $$\left$$($$\frac{2+x}{2}$$,\,$$\frac{3+y}{2}$$,\,$$\frac{5+z}{2}$$$$\right$$).$$

Equate each coordinate to the centre $$C(3,\,6,\,1)$$:

$$$$\frac{2+x}{2}$$=3 \;$$\Rightarrow$$\; x = 4$$
$$$$\frac{3+y}{2}$$=6 \;$$\Rightarrow$$\; y = 9$$
$$$$\frac{5+z}{2}$$=1 \;$$\Rightarrow$$\; z = -3$$

Therefore the other end of the diameter is $$Q(4,\,9,\,-3).$$

Option A which is: $$(4, 9, -3)$$