= Solution
{c}
The native PDF repeats $dW_y$ in its list of independent noise terms, whereas its third <stochastic differential equation> uses $dW_z$. We use the intended three independent coordinate <Brownian motions> $W_x,W_y,W_z$.
Put $\mathbf X=(X,Y,Z)$, $r=|\mathbf X|$ and $S=r^2$. Before the origin is reached, <Itô's formula> gives
$$
\begin{aligned}
dS&=2\mathbf X\mathbin{\cdot}d\mathbf X+(dX)^2+(dY)^2+(dZ)^2\\
&=(6-2\alpha)\,dt+2\sqrt2\,\mathbf X\mathbin{\cdot}d\mathbf W.
\end{aligned}
$$
The $6$ is the sum of three <quadratic variation> terms, each $2\,dt$. The <drift> contribution is $-2\alpha\,dt$ because the numerator is $|\mathbf X|^2$. The radial <stochastic integral> can be written using a scalar <Brownian motion> $B$:
$$
dS=2(3-\alpha)\,dt+2\sqrt{2S}\,dB,
\qquad
d(S/2)=(3-\alpha)\,dt+2\sqrt{S/2}\,dB.
$$
Thus $S/2$ is a <squared Bessel process> of dimension $\delta=3-\alpha$ until it reaches zero. If this equation is taken globally with the usual nonnegative <squared Bessel process> continuation, taking <expected values> gives
$$
\mathbb E S(t)=r_0^2+2(3-\alpha)t,
\qquad
\boxed{\lim_{t\to\infty}\frac{\mathbb E S(t)}t=6-2\alpha.}
$$
The <stochastic integral> has zero <expected value>: stopping first at large levels bounds the integrands, yields a finite first <moment> for $S$, and then makes $\mathbb E\int_0^tS(u)\,du$ finite, so the unrestricted integral is a square-integrable <martingale>.
\b[Boundary qualification.] The punctured domain in the question does not itself specify that continuation. The boundary classification of a <Bessel process> says that the origin is inaccessible from $r_0>0$ when $\delta\geq2$, equivalently $\alpha\leq1$. In this parameter range the original <stochastic differential equation> is global in the stated domain and the boxed result holds as written.
For $1<\alpha<3$, we have $0<\delta<2$: the origin is reached in finite time with <probability> one. One can also see accessibility directly from the radial <generator of an SDE diffusion>
$$
L_rf=f''+\frac{2-\alpha}{r}f'.
$$
For $r<b$, the <harmonic function> $1-(r/b)^{\alpha-1}$ is the <probability> of reaching zero before $b$. Finite-time access follows by using the nonnegative function
$$
v_b(r)=\frac{b^{3-\alpha}r^{\alpha-1}-r^2}{2(3-\alpha)},
\qquad L_rv_b=-1,\qquad v_b(0)=v_b(b)=0,
$$
in the stopped <Itô formula>; it bounds the <expected value> of the time to reach zero or $b$. Letting $b\to\infty$ gives almost sure finite hitting of zero. The coefficients of the original vector <stochastic differential equation> are undefined there, so its stated punctured-domain version has no prescribed long-time evolution in this range.
The natural continuing radial <squared Bessel process> gives $6-2\alpha$, but stopping and holding the particle at the origin would give a different answer. Indeed, with $T_0$ the first hit of zero, <Itô's formula> for the stopped process gives
$$
\mathbb E S(t\wedge T_0)=r_0^2+2(3-\alpha)\mathbb E(t\wedge T_0).
$$
Since $T_0<\infty$ almost surely, <dominated convergence> applied to $(t\wedge T_0)/t\leq1$ gives
$$
\lim_{t\to\infty}\frac{\mathbb E S(t\wedge T_0)}t=0.
$$
Therefore \b[$6-2\alpha$ is the usual continued-radial answer; for $1<\alpha<3$ a boundary convention is necessary.] It is not a uniquely defined limit for the punctured-domain model alone.
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