You essentially want to find the circumcenter $D$ of the triangle $\triangle_{ABC}$ with vertices
$$A = (-1,0,0), B = (0,2,0), C = (0,0,3)$$
Since $D$ is lying on the plane holding $A, B, C$, there exists $3$ real numbers
$\alpha,\beta,\gamma$ such that
$$D = \alpha A + \beta B + \gamma C\quad\text{ with }\quad \alpha + \beta + \gamma = 1$$
This $3$-tuple is called the baricenteric coordinate of $D$. They can be computed using the sides $a,b,c$ of the triangle alone:
$$\alpha : \beta : \gamma \;=\; a^2(-a^2 + b^2 + c^2) : b^2( a^2 - b^2 + c^2) : c^2(a^2 + b^2 - c^2)$$
For the triangle at hand, we have $(a^2,b^2,c^2) = (13,10,5)$. This leads to
$$\alpha : \beta : \gamma \;=\;
13(-13+10+5) : 10(13-10+5) : 5(13+10-5) = 26 : 80 : 90$$
As a result,
$\displaystyle\;D = \frac{26 A + 80 B + 90C}{26 + 80 + 90} = \left(-\frac{35}{98},\frac{40}{49},\frac{135}{98} \right).$
Hints:
(1) The given sphere is $\;(x+1)^2+(y-3)^2+z^2=9\;$
(2) The line through the above sphere's center and the given tangency point is $\;(-1,3,0)+t(1,2,-2)\;,\;\;t\in\Bbb R\;$
(3) In a similar way as in plane geometry, two tangent spheres' centers are joined by a line passing through the tangency point.
Best Answer
First, let's work out the plane of the circle.
The three triangle vertices are $$\vec{a} = \left [ \begin{array}{c} a \\ 0 \\ 0 \end{array} \right ], \qquad \vec{b} = \left [ \begin{array}{c} 0 \\ b \\ 0 \end{array} \right ], \qquad \vec{c} = \left [ \begin{array}{c} 0 \\ 0 \\ c \end{array} \right ] \tag{1}\label{na1}$$ The equation for the plane is $$\vec{p} \cdot \vec{n} - d = 0$$ where the plane normal $\vec{n}$ is $$\vec{n} = ( \vec{b} - \vec{a} ) \times ( \vec{c} - \vec{a} ) = \left [ \begin{array}{c} -a \\ b \\ 0 \end{array} \right ] \times \left [ \begin{array}{c} -a \\ 0 \\ c \end{array} \right ] = \left [ \begin{array}{c} b c \\ a c \\ a b \end{array} \right ]$$ and the distance from plane to origin $d$ is the same at all points on the plane, including the vertices, $$\vec{a} \cdot \vec{n} = \vec{b} \cdot \vec{n} = \vec{c} \cdot \vec{n} = a b c$$ i.e., the full equation of the plane is $$\vec{p} \cdot \vec{n} - a b c = 0, \qquad \vec{n} = \left [ \begin{array}{c} b c \\ a c \\ a b \end{array} \right ] \tag{2}\label{na2}$$
Now, one way to treat the circle is to treat it as a great circle of a sphere that passes through the points. The key point is that because it is a great circle, the center of the sphere must be on the same plane as the vertices.
This also means that we can use two coordinates, say $u$ and $v$, to describe the center of the circle (and sphere): $$\vec{p} = \vec{a} + u ( \vec{b} - \vec{a} ) + v (\vec{c} - \vec{a}) \tag{3}\label{na3}$$ If the radius of the circle (and therefore the radius of the sphere) is $r$, then we have a system of equations, $$\begin{cases} \left \lVert \vec{p} - \vec{a} \right \rVert^2 = r^2 \\ \left \lVert \vec{p} - \vec{b} \right \rVert^2 = r^2 \\ \left \lVert \vec{p} - \vec{c} \right \rVert^2 = r^2 \end{cases}$$ i.e. $$\begin{cases} \left \lVert u ( \vec{b} - \vec{a} ) + v (\vec{c} - \vec{a}) \right \rVert^2 - r^2 = 0 \\ \left \lVert u ( \vec{b} - \vec{a} ) + v (\vec{c} - \vec{a}) + \vec{a} - \vec{b} \right \rVert^2 - r^2 = 0 \\ \left \lVert u ( \vec{b} - \vec{a} ) + v (\vec{c} - \vec{a}) + \vec{a} - \vec{c} \right \rVert^2 - r^2 = 0 \end{cases} \tag{4}\label{na4}$$
This system of equations has two solutions, one of which has a negative radius $r$; thus only one solution that makes sense: $$\begin{cases} u = \frac{a^2 b^2 + b^2 c^2}{2 (a^2 b^2 + a^2 c^2 + b^2 c^2)} \\ v = \frac{a^2 c^2 + b^2 c^2}{2 (a^2 b^2 + a^2 c^2 + b^2 c^2)} \\ r = \sqrt{\frac{(a^2 + b^2)(a^2 + c^2)(b^2 + c^2)}{4 (a^2 b^2 + a^2 c^2 + b^2 c^2 )}} \tag{5}\label{na5}\end{cases}$$ To find the center $\vec{p}$ in Cartesian coordinates, insert $u$ and $v$ and $\eqref{na1}$ to $\eqref{na3}$.