If the plane $$2ax - 3ay + 4az + 6 = 0$$ passes through the midpoint of the line joining the centres of the spheres $$x^2 + y^2 + z^2 + 6x - 8y - 2z = 13$$ and $$x^2 + y^2 + z^2 - 10x + 4y - 2z = 8$$, then $$a$$ equals

The general equation of a sphere is $$x^{2}+y^{2}+z^{2}+2ux+2vy+2wz+d=0$$, whose centre is $$(-u,\,-v,\,-w)$$.

For the first sphere
$$x^{2}+y^{2}+z^{2}+6x-8y-2z=13$$,
compare with the general form:

$$2u=6\;\Rightarrow\;u=3,\qquad 2v=-8\;\Rightarrow\;v=-4,\qquad 2w=-2\;\Rightarrow\;w=-1.$$

Hence centre $$C_{1}=(-u,\,-v,\,-w)=(-3,\,4,\,1).$$

For the second sphere
$$x^{2}+y^{2}+z^{2}-10x+4y-2z=8$$,
we get:

$$2u=-10\;\Rightarrow\;u=-5,\qquad 2v=4\;\Rightarrow\;v=2,\qquad 2w=-2\;\Rightarrow\;w=-1.$$

Hence centre $$C_{2}=(-u,\,-v,\,-w)=(5,\,-2,\,1).$$

The midpoint $$M$$ of the line segment joining $$C_{1}$$ and $$C_{2}$$ is

$$M=\left(\frac{-3+5}{2},\,\frac{4+(-2)}{2},\,\frac{1+1}{2}\right)=(1,\,1,\,1).$$

The given plane is $$2ax-3ay+4az+6=0$$. Since the plane passes through $$M(1,1,1)$$, substitute these coordinates:

$$2a(1)-3a(1)+4a(1)+6=0$$ $$\Rightarrow\;(2a-3a+4a)+6=0$$ $$\Rightarrow\;3a+6=0$$ $$\Rightarrow\;a=-2.$$

Therefore the required value of $$a$$ is $$-2$$.

Option C which is: $$-2$$