= Solution
A <Schwartz function> is a smooth function for which every <seminorm>
$$
p_{\alpha\beta}(\varphi)=\sup_{x\in\mathbb R^n}|x^\alpha\partial^\beta\varphi(x)|
$$
is finite. These <seminorms> define the Fréchet topology of the <Schwartz space> $\mathcal S(\mathbb R^n)$. The <tempered distributions> form its continuous linear dual $\mathcal S'$. We use bilinear <distribution> pairing, with no complex conjugation on the <test function>.
Differentiation under the Fourier integral and <integration by parts> give
$$
\partial_\xi^\beta\widehat\varphi=\mathcal F[(-ix)^\beta\varphi],\qquad\xi^\alpha\widehat f=\mathcal F[D^\alpha f],\quad D=-i\partial.
$$
Therefore each <Schwartz seminorm> of $\widehat\varphi$ is bounded by a finite sum of $L^1$ norms of polynomially weighted <derivatives> of $\varphi$. Insert the integrable weight $\langle x\rangle^{-n-1}$ to bound those norms by finitely many <Schwartz seminorms>. This proves that $\mathcal F:\mathcal S\to\mathcal S$ is continuous.
To justify inversion without a merely formal exchange of integrals, insert $e^{-\varepsilon|\xi|^2/2}$ in the inverse integral. The Gaussian Fourier integral gives
$$
(2\pi)^{-n}\int e^{ix\cdot\xi}e^{-\varepsilon|\xi|^2/2}\widehat\varphi(\xi)d\xi=(\varphi*g_\varepsilon)(x),\qquad g_\varepsilon(x)=(2\pi\varepsilon)^{-n/2}e^{-|x|^2/(2\varepsilon)}.
$$
The normalized Gaussian is an approximate identity, so the right side tends to $\varphi(x)$; the left side converges by <dominated convergence theorem> since $\widehat\varphi$ is integrable. It follows that
$$
\boxed{\varphi(x)=(2\pi)^{-n}\int e^{ix\cdot\xi}\widehat\varphi(\xi)d\xi,\qquad\mathcal F^2\varphi=(2\pi)^n\varphi(-\cdot).}
$$
Reflection is continuous in <Schwartz seminorms>, so $\mathcal F^{-1}=(2\pi)^{-n}\mathcal R\mathcal F$ is continuous as well. This proves the <Fourier transform isomorphism of the Schwartz space>.
Define the <distributional Fourier transform> by
$$
\boxed{\langle\widehat u,\varphi\rangle=\langle u,\widehat\varphi\rangle.}
$$
It agrees with the ordinary integral transform whenever Fubini applies. Transposing the Schwartz inverse gives an inverse on $\mathcal S'$, and the same squared-transform identity holds. For the usual strong dual topology, a <seminorm> is $p_B(u)=\sup_{\varphi\in B}|\langle u,\varphi\rangle|$ for a bounded set $B\subset\mathcal S$. Since a continuous linear Schwartz map takes bounded sets to bounded sets, $p_B(\widehat u)=p_{\mathcal F B}(u)$ proves <continuity>, and similarly for the inverse. <Continuity> also holds in the weak dual topology. Thus the transform is a continuous isomorphism on <tempered distributions>, with the stated normalization.
For a real symmetric <positive-definite matrix> $G$, there is $c>0$ with $g(x)=x^TGx\geq c|x|^2$. The reciprocal is locally integrable in dimension three: the radial factor near zero is $r^2r^{-2}dr$. At infinity, rapid decay of a Schwartz <test function> makes it integrable. More quantitatively,
$$
\left|\int\frac{\varphi(x)}{g(x)}dx\right|\leq C\sup_x\langle x\rangle^2|\varphi(x)|\int_{\mathbb R^3}\frac{dx}{|x|^2\langle x\rangle^2}<\infty.
$$
This single <seminorm> bound proves a <tempered distribution>. No principal-value extension is needed at the origin.
First calculate the isotropic transform. For $f_\varepsilon(x)=e^{-\varepsilon|x|}/|x|^2$ and $k=|\xi|>0$, spherical integration and the supplied sine-integral identity give
$$
\widehat f_\varepsilon(\xi)=\frac{4\pi}{k}\int_0^\infty e^{-\varepsilon r}\frac{\sin(kr)}rdr=\frac{4\pi}{k}\arctan\frac k\varepsilon.
$$
The original functions converge in $\mathcal S'$ to $|x|^{-2}$ by <dominated convergence theorem> against tests. The transforms are bounded by $2\pi^2/|\xi|$, which is locally integrable in three dimensions and integrable against <Schwartz functions> at infinity. Hence
$$
\boxed{\mathcal F(|x|^{-2})(\xi)=\frac{2\pi^2}{|\xi|}\quad\text{as a regular tempered distribution}.}
$$
Let $y=G^{1/2}x$. The Jacobian is $(\det G)^{-1/2}$ and the dual vector is $G^{-1/2}\xi$. The <Fourier transform of a reciprocal positive quadratic form> is consequently
$$
\boxed{\widehat{(1/g)}(\xi)=\frac{2\pi^2}{\sqrt{\det G}\sqrt{\xi^TG^{-1}\xi}}.}
$$
The frequency-origin value is understood distributionally; there is no additional delta term.
For the complex extension use a symmetric <matrix> $A=G+iB$ with real symmetric $B$ and positive real part $G$. Symmetry is natural for a <quadratic form>; a skew-symmetric part contributes nothing. The bound $|x^TAx|\geq x^TGx$ again gives a regular tempered reciprocal. Both sides of the prospective formula depend holomorphically on the <matrix> while its real part is positive: compact parameter sets give a common $|x|^{-2}$ bound for pairing and differentiated integrands. Continue from the real positive <matrices> along $A_z=G+izB$. In a complex neighborhood of $z=0$, imaginary $z$ gives real positive <matrices>, so the one-variable identity theorem supplies the equality; connected continuation along $0\leq z\leq1$ reaches $A$.
The <analytic determinant square root for accretive symmetric matrices> must follow that continuation, rather than an arbitrary scalar principal root of $\det A$. Explicitly, put $C=G^{-1/2}BG^{-1/2}$ and let its real <eigenvalues> be $b_j$. Then
$$
d(A)=\sqrt{\det G}\prod_{j=1}^3\sqrt{1+ib_j},
$$
where each factor has positive real part. This is the <determinant> square root normalized positively on real positive <matrices>. Also
$$
\operatorname{Re}(\xi^TA^{-1}\xi)=\xi^TG^{-1/2}(1+C^2)^{-1}G^{-1/2}\xi>0\qquad(\xi\ne0).
$$
Thus the quadratic-form square root has the unambiguous branch with positive real part. The <accretive complex quadratic reciprocal Fourier transform> is
$$
\boxed{\mathcal F\!\left(\frac1{x^TAx}\right)(\xi)=\frac{2\pi^2}{d(A)\sqrt{\xi^TA^{-1}\xi}}.}
$$
Uniform local integrability of the right side justifies its <analytic continuation> as a <tempered distribution>, not merely pointwise away from the origin.
For the particular form,
$$
A=\begin{pmatrix}1&i&0\\i&1&0\\0&0&2\end{pmatrix},\qquad\det A=4,\qquad d(A)=2,\qquad A^{-1}=\frac12\begin{pmatrix}1&-i&0\\-i&1&0\\0&0&1\end{pmatrix}.
$$
Substitution gives the required normalized expression
$$
\boxed{\mathcal F\!\left(\frac1{x_1^2+x_2^2+2x_3^2+2ix_1x_2}\right)(\xi)=\frac{\sqrt2\pi^2}{\sqrt{\xi_1^2+\xi_2^2+\xi_3^2-2i\xi_1\xi_2}}.}
$$
For every nonzero real $\xi$, the radicand has positive real part, so the specified square root exists uniquely. Both sides are regular <tempered distributions>, despite their locally integrable singularities at zero.
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