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The equation of a sphere is $(x-x_0) ^2 + (y - y_0)^2 + (z-z_0)^2 = r^2$ where the center of the sphere is $(x_0, y_0, z_0)$. This should give you three equations (one for each point) with three unknowns. You can then solve for the unknowns. Can you take it from here?
Bob
On
Sketch of an algorithm, too long for a comment. Maybe someone will edit to provide the actual linear algebra.
Translate the three points by $-P2$ to move $P2$ to the origin.
Rotate to move (the translates of) $P1$ and $P3$ to the $x-y$ plane.
Solve the two dimensional problem.
Rotate and translate the answer back.
On
This is a 2-D problem in disguise, so let’s solve it there first. Assuming that $P_2$ is the vertex, the center of the arc must lie somewhere on the angle bisector of $\angle{P_1P_2P_3}$. Let $\mathbf u_{21}={P_1-P_2\over\|P_1-P_2\|}$ and $\mathbf u_{23}={P_3-P_2\over\|P_3-P_2\|}$, the unit vectors in the directions of the two given rays. Then $\mathbf v=\mathbf u_{21}+\mathbf u_{23}$ is their angle bisector, and so $P_c=P_2+\lambda\mathbf v$. The arc has a given radius $r$, so we just need to find the value of $\lambda$ for which the distance of $P_c$ from either of the lines $\overline{P_2P_1}$ or $\overline{P_2P_3}$ is equal to $r$. This distance is equal to the length of the orthogonal rejection of $\lambda\mathbf v$ from either $\mathbf u_{21}$ or $\mathbf u_{23}$, i.e., $$\|\lambda\mathbf v-(\lambda\mathbf v\cdot\mathbf u)\mathbf u\|=r.$$ By construction $\lambda\ge0$, so this gives $$\lambda = {r\over\|\mathbf v-(\mathbf v\cdot\mathbf u)\mathbf u\|}.$$ With $\lambda$ in hand, we can compute the three points of interest: $$P_c=P_2+\lambda\mathbf v \\ P_a = P_2+\lambda(\mathbf v\cdot\mathbf u_{21})\mathbf u_{21} \\ P_b = P_2+\lambda(\mathbf v\cdot\mathbf u_{23})\mathbf u_{23}.$$ (The last two are just the projections of $P_c$ onto the two given rays.)
Nothing in the above solution relies on the points and vectors being two-dimensional, so it transfers directly to 3-D. Depending on how precise your values for the initial points are, you might get a better result in practice by computing $\lambda$ using each of the two unit vectors and taking the average of the two resulting values.
It may help with the computations to note that $$\mathbf v\cdot\mathbf u_{21}=\mathbf v\cdot\mathbf u_{23}=1+\mathbf u_{21}\cdot\mathbf u_{23}$$ and so $$\mathbf v-(\mathbf v\cdot\mathbf u_{23})\mathbf u_{23} = \mathbf u_{21}-(\mathbf u_{21}\cdot\mathbf u_{23})\mathbf u_{23} \\ \mathbf v-(\mathbf v\cdot\mathbf u_{21})\mathbf u_{21} = \mathbf u_{23}-(\mathbf u_{21}\cdot\mathbf u_{23})\mathbf u_{21}$$ that is, the orthogonal rejection of $\mathbf v$ from one of the unit vectors is equal to the orthogonal rejection of the other unit vector from it (which, by symmetry, all have equal lengths). Thus, computing $\mathbf v$ explicitly might not be necessary (although it’s only a couple of additions). However, I don’t see a good way to avoid having to compute three square roots when following this approach.
In fact, it's a 2D problem, you just have to choose the right coordinate system. Let $$\vec{P_2P_1}=P_1-P_2$$ the vector pointing from $P_2$ to $P_1,$ and $$n_a=\frac{\vec{P_2P_1}}{|\vec{P_2P_1}|}$$ the normalized (length $1$) vector, $$\vec{P_2P_3}=P_3-P_2$$ the vector pointing from $P_2$ to $P_3,$ and $$n_b=\frac{\vec{P_2P_3}}{|\vec{P_2P_3}|}$$ the normalized vector. The angle $\alpha$ between these two directions is defined by the scalar product: $$\cos\alpha=n_a\cdot n_b.$$ Clearly, $P_c$ lies on the bisector of this angle. The distance from $P_2$ to $P_a$ and $P_b$ must be the same, $r/\tan\frac{\alpha}{2}$, so $$P_a=P_2+\frac{r}{\tan\frac{\alpha}{2}}\,n_a$$ and $$P_b=P_2+\frac{r}{\tan\frac{\alpha}{2}}\,n_b,$$ and you get $P_c$ from $$P_c=P_2+\frac{r}{\sin\alpha}\,(n_a+n_b).$$
While these formulas show the simple geometry more clearly, @Nominal Animal pointed out that it might be better to avoid calculating trigonometric functions. Indeed, we can use the cross product to express $\sin\alpha$: $$\sin\alpha=|n_a\times n_b|.$$ For calculating $P_a, P_b$, we can use $$\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha}=\frac{|n_a\times n_b|}{1+n_a\cdot n_b}.$$ So everything can be determined by vector operations, alone.