= Solution
For a <decision rule> $\delta$ and loss $L$, its <risk function> is $R(\theta,\delta)=\mathbb E_\theta L(\theta,\delta(X))$. It is an <admissible estimator> if there is no rule with risk everywhere no larger and somewhere strictly smaller. A <Bayes decision rule> minimizes $\int R(\theta,\delta)\,\pi(d\theta)$ for a specified prior $\pi$. A <minimax decision rule> minimizes $\sup_\theta R(\theta,\delta)$ over all rules.
Here the <maximum-likelihood estimator> is $\delta_0(X)=X$, and its squared-error risk is
$$
\boxed{R(\theta,\delta_0)=\sum_{i=1}^3\mathbb E_\theta(X_i-\theta_i)^2=3.}
$$
Write the proposed estimator as $\delta(X)=X+g(X)$ with $g(x)=-x/\|x\|^2$ for $x\ne0$, and assign any value at zero, an event of probability zero. Expanding the squared loss gives
$$
R(\theta,\delta)=3+\mathbb E_\theta\|g(X)\|^2+2\sum_i\mathbb E_\theta[(X_i-\theta_i)g_i(X)].
$$
<Stein's lemma> for a unit-covariance normal vector says $\mathbb E[(X_i-\theta_i)g_i(X)]=\mathbb E[\partial_i g_i(X)]$ when the relevant derivatives and expectations are integrable. With $r=\|x\|$,
$$
\|g(x)\|^2=r^{-2},\qquad\sum_{i=1}^3\partial_ig_i(x)=-\frac3{r^2}+\frac{2\sum_i x_i^2}{r^4}=-r^{-2}.
$$
The singularity causes no difficulty in three dimensions: $r^{-2}$ is locally integrable against the volume element $r^2dr\,d\Omega$, and the normal tails control infinity. More formally apply Stein's lemma first to $g_\epsilon(x)=-x/(r^2+\epsilon)$, then let $\epsilon\downarrow0$; its squared norm and derivative magnitudes are bounded by constants times $r^{-2}$, so <dominated convergence> justifies the limit. Consequently
$$
\boxed{R(\theta,\delta)=3-\mathbb E_\theta\frac1{\|X\|^2}<3\quad\text{for every }\theta\in\mathbb R^3.}
$$
The expectation is finite and strictly positive. Thus this <James–Stein estimator> strictly dominates the maximum likelihood estimator, which is inadmissible under the specified squared-error loss.
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