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For three non-collinear points with position vectors $\mathbf{a},\mathbf{b},\mathbf{c}$, a nonzero normal vector isThe plane is the set of $\mathbf{r}$ satisfying $(\mathbf{r}-\mathbf{a})\cdot \mathbf{N}=0$. Expanding the cross product and using the scalar triple product givesNon-collinearity is essential: otherwise $\mathbf{N}=0$ and the equation fails to select a unique plane.

For a point $\mathbf{x}$ on the plane, the Cauchy-Schwarz inequality gives $|d|=|\mathbf{x}\cdot \mathbf{n}|\le |\mathbf{x}|$, because $\mathbf{n}$ is a unit vector. Equality is achieved at $\mathbf{x}=d\mathbf{n}$. Thus 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 givesIf $|\delta |=r$, the sphere equation therefore forces $\mathbf{u}=0$. 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 vectorsTheir 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 givesSolving proves the center formulaThis 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 $\mathbf{N}\ne 0$, and the plane through the three vertices isThe second formula follows by normalizing $\mathbf{N}$ to a unit vector and applying the first distance result.

Use suffix notation with indices ranging from $1$ to $3$, and sum over repeated indices. By the component definitions of the dot product and cross product,The Levi-Civita symbol is unchanged by a cyclic permutation of its three indices: the cycle is two transpositions and hence has positive sign. Thus ${ϵ}_{ijk}={ϵ}_{kij}$. Commuting the scalar components givesTherefore the scalar triple product obeysCyclically moving the vectors preserves the scalar triple product; interchanging just two reverses its sign.

The two displacement vectors $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$ lie in the required plane. Because the three points are non-collinear, their cross productis nonzero and is a normal vector to that plane. A point with position vector $\mathbf{r}$ lies in the plane exactly whenExpanding the cross product givesThe terms $\mathbf{a}\cdot (\mathbf{a}\times \mathbf{b})$ and $\mathbf{a}\cdot (\mathbf{c}\times \mathbf{a})$ vanish, so the equation of a plane through three points becomesNon-collinearity is important: it ensures this is a genuine plane equation with nonzero normal, rather than the vacuous equality $0=0$.

For the specified coordinates,Their sum is $(3,3,6)$, with Euclidean norm $3\sqrt{6}$. Consequently either orientation of the unit normal is valid:As a check, the scalar triple product is $12$, so the plane equation simplifies to $x+y+2z=4$, which all three supplied points satisfy. The normal direction is $(1,1,2)$.