= Solution
For a point $\mathbf x$ on the <plane>, the <Cauchy-Schwarz inequality> gives $|d|=|\mathbf x\cdot\mathbf n|\leq|\mathbf x|$, because $\mathbf n$ is a <unit vector>. Equality is achieved at $\mathbf x=d\mathbf n$. Thus \b[the distance from the origin is $|d|$].
Let $\delta=\mathbf p\cdot\mathbf n-d$ and let $\mathbf q=\mathbf p-\delta\mathbf n$ be the perpendicular projection of the <sphere>'s center onto the <plane>. Every point of the <plane> has form $\mathbf x=\mathbf q+\mathbf u$ with $\mathbf u\cdot\mathbf n=0$. The <Pythagorean theorem> gives
$$
|\mathbf x-\mathbf p|^2=|\mathbf u-\delta\mathbf n|^2=|\mathbf u|^2+\delta^2.
$$
If $|\delta|=r$, the <sphere> equation therefore forces $\mathbf u=0$. \b[There is exactly one contact point, $\mathbf q$]; it lies on both surfaces.
Write $\tau=\mathbf a\cdot(\mathbf b\times\mathbf c)>0$. The nonzero <scalar triple product> proves that $\mathbf a,\mathbf b,\mathbf c$ are <linearly independent>, so they form a positively oriented <basis> of $\mathbb R^3$ and define a nondegenerate <tetrahedron>.
For the three faces through the origin, choose the inward <unit vectors>
$$
\mathbf n_a=\frac{\mathbf b\times\mathbf c}{|\mathbf b\times\mathbf c|},\quad
\mathbf n_b=\frac{\mathbf c\times\mathbf a}{|\mathbf c\times\mathbf a|},\quad
\mathbf n_c=\frac{\mathbf a\times\mathbf b}{|\mathbf a\times\mathbf b|}.
$$
Their signs are correct because each has positive <inner product> with the opposite vertex vector. Write the center as $\mathbf p=\alpha\mathbf a+\beta\mathbf b+\gamma\mathbf c$. It lies on the interior side of every face. Tangency of the interior <sphere> therefore means signed distance $r$, not $-r$, from each of those three <planes>. Taking <inner products> gives
$$
\alpha\frac{\tau}{|\mathbf b\times\mathbf c|}=r,\quad
\beta\frac{\tau}{|\mathbf c\times\mathbf a|}=r,\quad
\gamma\frac{\tau}{|\mathbf a\times\mathbf b|}=r.
$$
Solving proves \b[the center formula]
$$
\boxed{\mathbf p=\frac r\tau\left(|\mathbf b\times\mathbf c|\mathbf a+|\mathbf c\times\mathbf a|\mathbf b+|\mathbf a\times\mathbf b|\mathbf c\right).}
$$
This is the <sphere center from three tetrahedron face distances>; the interior hypothesis selects these signs rather than an exterior center.
Finally set $\mathbf N=\mathbf a\times\mathbf b+\mathbf b\times\mathbf c+\mathbf c\times\mathbf a$. The <scalar triple product> identities give $\mathbf N\cdot\mathbf a=\mathbf N\cdot\mathbf b=\mathbf N\cdot\mathbf c=\tau$. In particular $\mathbf N\ne0$, and the <plane> through the three vertices is
$$
\boxed{\Psi:\ \mathbf N\cdot\mathbf x=\tau,\qquad \operatorname{dist}(O,\Psi)=\frac{\tau}{|\mathbf N|}.}
$$
The second formula follows by normalizing $\mathbf N$ to a <unit vector> and applying the first distance result.
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