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The locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_2$ is a sheet of a hyperboloid, whose equation can be found from $PC_2-PC_1=b-a$. Using the formulas for $C_1$, $C_2$, $C_3$ given in the question, this equation can be expanded to: $$ 4 (a-b)^2 \left(\left(x-2 \sqrt{a b}\right)^2+(b-z)^2+y^2\right)=4 \left(-2 x \sqrt{a b}+a (b+z)+b^2-b z\right)^2. $$ Analogously, the locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_3$ is a sheet of another hyperboloid, with equation: $$ 4 (a-c)^2 \left(\left(\frac{b c-a (b+c)}{\sqrt{a b}}+x\right)^2+\left(\sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-y\right)^2+(c-z)^2\right)=4 \left(-c \left(\frac{a x}{\sqrt{a b}}+z\right)+\frac{b c x}{\sqrt{a b}}-y \sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-x \sqrt{a b}+a (c+z)+c^2\right)^2. $$ The intersection of those two hyperboloids gives then the locus of the center of the fourth sphere, with center $P$ and tangent to the other three spheres. The limit position of $P$, to avoid the fourth sphere to fall, is on the plane $y=0$, which is the plane passing through the centers of the two smallest spheres and perpendicular to $(xy)$ plane. Hence this position of $P$ can be found by solving the system formed by the two above equations and $y=0$.

Unfortunately, the resultant equation is a cubic, leading to three solutions quite long and involved to write down. For $𝑎=15$, $𝑏=20$, $𝑐=25$ I get for instance (with the help of Mathematica): $$ \begin{align} P_1&=\left({180\sqrt3\over41},0,{45480\over943}\right)\\ P_2&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889+37 \sqrt{409}\right)} ,0,-\frac{25}{64} \left(119+11 \sqrt{409}\right)\right)\\ P_3&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889-37 \sqrt{409}\right)} ,0,\frac{25}{64} \left(-119+11 \sqrt{409}\right)\right)\\ \end{align} $$ But only the first solution corresponds to a point lying on the "right" sheet of the hyperboloids, i.e. $P_1$ is the only position of $P$ satisfying $PC_1-a=PC_2-b=PC_3-c=r$. The corresponding radius is: $$ r=\frac{18000}{943}. $$ This is then the maximum radius of the fourth sphere.

The locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_2$ is a sheet of a hyperboloid, whose equation can be found from $PC_2-PC_1=b-a$. Using the formulas for $C_1$, $C_2$, $C_3$ given in the question, this equation can be expanded to: $$ 4 (a-b)^2 \left(\left(x-2 \sqrt{a b}\right)^2+(b-z)^2+y^2\right)=4 \left(-2 x \sqrt{a b}+a (b+z)+b^2-b z\right)^2. $$ Analogously, the locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_3$ is a sheet of another hyperboloid, with equation: $$ 4 (a-c)^2 \left(\left(\frac{b c-a (b+c)}{\sqrt{a b}}+x\right)^2+\left(\sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-y\right)^2+(c-z)^2\right)=4 \left(-c \left(\frac{a x}{\sqrt{a b}}+z\right)+\frac{b c x}{\sqrt{a b}}-y \sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-x \sqrt{a b}+a (c+z)+c^2\right)^2. $$ The intersection of those two hyperboloids gives then the locus of the center of the fourth sphere, with center $P$ and tangent to the other three spheres. The limit position of $P$, to avoid the fourth sphere to fall, is on the plane $y=0$, which is the plane passing through the centers of the two smallest spheres and perpendicular to $(xy)$ plane. Hence this position of $P$ can be found by solving the system formed by the two above equations and $y=0$.

This can also be seen as the intersection of two hyperbolas (shown in figure below), whose equations in the $(xz)$ plane are obtained by plugging $y=0$ in the above equations.

Unfortunately, the resultant equation is a cubic, leading to three solutions quite long and involved to write down in their general form, as a function of $a$, $b$ and $c$. For $𝑎=15$, $𝑏=20$, $𝑐=25$ I got (with the help of Mathematica): $$ \begin{align} P_1&=\left({180\sqrt3\over41},0,{45480\over943}\right)\\ P_2&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889+37 \sqrt{409}\right)} ,0,-\frac{25}{64} \left(119+11 \sqrt{409}\right)\right)\\ P_3&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889-37 \sqrt{409}\right)} ,0,\frac{25}{64} \left(-119+11 \sqrt{409}\right)\right)\\ \end{align} $$ But only the first solution corresponds to a point lying on the "right" sheet of the hyperboloids, i.e. $P_1$ is the only position of $P$ satisfying $PC_1-a=PC_2-b=PC_3-c=r$. The corresponding radius is: $$ r=\frac{18000}{943}. $$ This is then the maximum radius of the fourth sphere.

The locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_2$ is a sheet of a hyperboloid, whose equation can be found from $PC_2-PC_1=b-a$ and can be developed to: $$ 4 (a-b)^2 \left(\left(x-2 \sqrt{a b}\right)^2+(b-z)^2+y^2\right)=4 \left(-2 x \sqrt{a b}+a (b+z)+b^2-b z\right)^2. $$ Analogously, the locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_3$ is a sheet of another hyperboloid, with equation: $$ 4 (a-c)^2 \left(\left(\frac{b c-a (b+c)}{\sqrt{a b}}+x\right)^2+\left(\sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-y\right)^2+(c-z)^2\right)=4 \left(-c \left(\frac{a x}{\sqrt{a b}}+z\right)+\frac{b c x}{\sqrt{a b}}-y \sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-x \sqrt{a b}+a (c+z)+c^2\right)^2. $$ The intersection of those two hyperboloids gives then the locus of the center of the fourth sphere, with center $P$ and tangent to the other three spheres. The limit position of $P$, to avoid the fourth sphere to fall, is on the plane $y=0$, which is the plane passing through the centers of the two smallest spheres and perpendicular to $(xy)$ plane. Hence this position of $P$ can be found by solving the system formed by the two above equations and $y=0$.

Unfortunately, the resultant equation is a cubic, leading to three solutions quite long and involved to write down. For $𝑎=15$, $𝑏=20$, $𝑐=25$ I get for instance (with the help of Mathematica): $$ \begin{align} P_1&=\left({180\sqrt3\over41},0,{45480\over943}\right)\\ P_2&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889+37 \sqrt{409}\right)} ,0,-\frac{25}{64} \left(119+11 \sqrt{409}\right)\right)\\ P_3&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889-37 \sqrt{409}\right)} ,0,\frac{25}{64} \left(-119+11 \sqrt{409}\right)\right)\\ \end{align} $$ But only the first solution corresponds to a point lying on the "right" sheet of the hyperboloids. The corresponding radius is: $$ r=\frac{18000}{943}. $$ This is then the maximum radius of the fourth sphere.

The locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_2$ is a sheet of a hyperboloid, whose equation can be found from $PC_2-PC_1=b-a$. Using the formulas for $C_1$, $C_2$, $C_3$ given in the question, this equation can be expanded to: $$ 4 (a-b)^2 \left(\left(x-2 \sqrt{a b}\right)^2+(b-z)^2+y^2\right)=4 \left(-2 x \sqrt{a b}+a (b+z)+b^2-b z\right)^2. $$ Analogously, the locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_3$ is a sheet of another hyperboloid, with equation: $$ 4 (a-c)^2 \left(\left(\frac{b c-a (b+c)}{\sqrt{a b}}+x\right)^2+\left(\sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-y\right)^2+(c-z)^2\right)=4 \left(-c \left(\frac{a x}{\sqrt{a b}}+z\right)+\frac{b c x}{\sqrt{a b}}-y \sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-x \sqrt{a b}+a (c+z)+c^2\right)^2. $$ The intersection of those two hyperboloids gives then the locus of the center of the fourth sphere, with center $P$ and tangent to the other three spheres. The limit position of $P$, to avoid the fourth sphere to fall, is on the plane $y=0$, which is the plane passing through the centers of the two smallest spheres and perpendicular to $(xy)$ plane. Hence this position of $P$ can be found by solving the system formed by the two above equations and $y=0$.

Unfortunately, the resultant equation is a cubic, leading to three solutions quite long and involved to write down. For $𝑎=15$, $𝑏=20$, $𝑐=25$ I get for instance (with the help of Mathematica): $$ \begin{align} P_1&=\left({180\sqrt3\over41},0,{45480\over943}\right)\\ P_2&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889+37 \sqrt{409}\right)} ,0,-\frac{25}{64} \left(119+11 \sqrt{409}\right)\right)\\ P_3&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889-37 \sqrt{409}\right)} ,0,\frac{25}{64} \left(-119+11 \sqrt{409}\right)\right)\\ \end{align} $$ But only the first solution corresponds to a point lying on the "right" sheet of the hyperboloids, i.e. $P_1$ is the only position of $P$ satisfying $PC_1-a=PC_2-b=PC_3-c=r$. The corresponding radius is: $$ r=\frac{18000}{943}. $$ This is then the maximum radius of the fourth sphere.

The locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_2$ is a sheet of a hyperboloid, whose equation can be found from $PC_2-PC_1=b-a$ and can be developed to: $$ 4 (a-b)^2 \left(\left(x-2 \sqrt{a b}\right)^2+(b-z)^2+y^2\right)=4 \left(-2 x \sqrt{a b}+a (b+z)+b^2-b z\right)^2. $$ Analogously, the locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_3$ is a sheet of another hyperboloid, with equation: $$ 4 (a-c)^2 \left(\left(\frac{b c-a (b+c)}{\sqrt{a b}}+x\right)^2+\left(\sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-y\right)^2+(c-z)^2\right)=4 \left(-c \left(\frac{a x}{\sqrt{a b}}+z\right)+\frac{b c x}{\sqrt{a b}}-y \sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-x \sqrt{a b}+a (c+z)+c^2\right)^2. $$ The intersection of those two hyperboloids gives then the locus of the center of the fourth sphere, with center $P$ and tangent to the other three spheres. The limit position of $P$, to avoid the fourth sphere to fall, is on the plane $y=0$, which is the plane passing through the centers of the two smallest spheres and perpendicular to $(xy)$ plane. Hence this position of $P$ can be found by solving the system formed by the two above equations and $y=0$.

Unfortunately, the resultant equation is a cubic, leading to three solutions quite long and involved to write down. For $𝑎=15$, $𝑏=20$, $𝑐=25$ I get for instance (with the help of Mathematica): $$ \begin{align} P_1&=\left({180\sqrt3\over41},0,{45480\over943}\right)\\ P_2&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889+37 \sqrt{409}\right)} ,0,-\frac{25}{64} \left(119+11 \sqrt{409}\right)\right)\\ P_3&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889-37 \sqrt{409}\right)} ,0,\frac{25}{64} \left(-119+11 \sqrt{409}\right)\right)\\ \end{align} $$ But only the first solution corresponds to a point lying on the "right" sheet of the hyperboloids. The corresponding radius is: $$ r=\frac{18000}{943}. $$ This is then the maximum radius of the fourth sphere. The minimun has been found in the question.

The locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_2$ is a sheet of a hyperboloid, whose equation can be found from $PC_2-PC_1=b-a$ and can be developed to: $$ 4 (a-b)^2 \left(\left(x-2 \sqrt{a b}\right)^2+(b-z)^2+y^2\right)=4 \left(-2 x \sqrt{a b}+a (b+z)+b^2-b z\right)^2. $$ Analogously, the locus of the center $P=(x,y,z)$ of a sphere tangent to spheres $C_1$ and $C_3$ is a sheet of another hyperboloid, with equation: $$ 4 (a-c)^2 \left(\left(\frac{b c-a (b+c)}{\sqrt{a b}}+x\right)^2+\left(\sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-y\right)^2+(c-z)^2\right)=4 \left(-c \left(\frac{a x}{\sqrt{a b}}+z\right)+\frac{b c x}{\sqrt{a b}}-y \sqrt{4 a c-\frac{(b c-a (b+c))^2}{a b}}-x \sqrt{a b}+a (c+z)+c^2\right)^2. $$ The intersection of those two hyperboloids gives then the locus of the center of the fourth sphere, with center $P$ and tangent to the other three spheres. The limit position of $P$, to avoid the fourth sphere to fall, is on the plane $y=0$, which is the plane passing through the centers of the two smallest spheres and perpendicular to $(xy)$ plane. Hence this position of $P$ can be found by solving the system formed by the two above equations and $y=0$.

Unfortunately, the resultant equation is a cubic, leading to three solutions quite long and involved to write down. For $𝑎=15$, $𝑏=20$, $𝑐=25$ I get for instance (with the help of Mathematica): $$ \begin{align} P_1&=\left({180\sqrt3\over41},0,{45480\over943}\right)\\ P_2&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889+37 \sqrt{409}\right)} ,0,-\frac{25}{64} \left(119+11 \sqrt{409}\right)\right)\\ P_3&=\left(\frac{5}{4} \sqrt{\frac{3}{2} \left(889-37 \sqrt{409}\right)} ,0,\frac{25}{64} \left(-119+11 \sqrt{409}\right)\right)\\ \end{align} $$ But only the first solution corresponds to a point lying on the "right" sheet of the hyperboloids. The corresponding radius is: $$ r=\frac{18000}{943}. $$ This is then the maximum radius of the fourth sphere.