Explanation

(b)
Let the equation of the sphere be $x^{2} + y^{2} + \mathcal{z}^{2} - ax - by - cz = 0$. This meets the axes at $A(a,\ 0,\ 0),\ B(0,\ b,\ 0)$ and $C(0,\ 0,\ c)$
Let $(\alpha,\ \beta,\ \gamma)$ be the coordinares of the centroid of the tetrahedron $OABC$. Then
$\frac{a}{4} = \alpha,\frac{b}{4} = \beta,\frac{c}{4} = \gamma$
$\Rightarrow a = 4\alpha,\ b = 4\beta,\ c = 4\gamma$
Now, radius of the sphere $= 2k$
$\Rightarrow \frac{1}{2}\sqrt{a^{2} + b^{2} + c^{2}} = 2k\ \Rightarrow a^{2} + b^{2} + c^{2} = 16k^{2}$
$\Rightarrow 16\left( \alpha^{2} + \beta^{2} + \gamma^{2} \right) = 16k^{2}$
Hence, the locus of $(\alpha,\ \beta,\ \gamma)$ is $\left( x^{2} + y^{2} + \mathcal{z}^{2} \right) = k^{2}$