Midpoints of the spheres $M_{1}(0,-2,0),M_{2}(0,2,0),M_{3}(\sqrt{21},0,0),M_{4}(\frac{\sqrt{21}}{7},0,\frac{12\sqrt{7}}{7})$, see two pictures. $M_{1}$ green, $M_{2}$ red, $M_{3}$ blue, $M_{4}$ purple. So $M_{1}M_{2}=4,M_{1}M_{3}=5,M_{1}M_{4}=5,M_{2}M_{3}=5,M_{2}M_{4}=5,M_{3}M_{4}=6$. Tetrahedron $(M_{4},M_{1}M_{2}M_{3})$. Midpoint $M(c,0,d)$ and radius $r$, satisfying: \[\left\{\begin{array}{l} MM_{2}+2=r \\ MM_{3}+3=r \\ MM_{4}+3=r \\ \end{array} \right.\] \[\left\{\begin{array}{l} \sqrt{c^{2}+4+d^{2}}+2=r \\ \sqrt{(c-\sqrt{21})^{2}+d^{2}}+3=r \\ \sqrt{(c-\frac{\sqrt{21}}{7})^{2}+(d-\frac{12\sqrt{7}}{7})^{2}}+3=r \\ \end{array} \right.\] Solving $M(\frac{4\sqrt{21}}{7},0,\frac{6\sqrt{7}}{7})$ and $r=6$.

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vanstraelen wrote: Midpoints of the spheres $M_{1}(0,-2,0),M_{2}(0,2,0),M_{3}(\sqrt{21},0,0),M_{4}(\frac{\sqrt{21}}{7},0,\frac{12\sqrt{7}}{7})$, see two pictures. $M_{1}$ green, $M_{2}$ red, $M_{3}$ blue, $M_{4}$ purple. So $M_{1}M_{2}=4,M_{1}M_{3}=4,M_{1}M_{4}=5,M_{2}M_{3}=5,M_{2}M_{4}=5,M_{3}M_{4}=6$. Tetrahedron $(M_{4},M_{1}M_{2}M_{3}$. Modpoint $M(c,0,d)$ and radius $r$, satisfying: \[\left\{\begin{array}{l} MM_{2}+2=r \\ MM_{3}+3=r \\ MM_{4}+3=r \\ \end{array} \right.\] \[\left\{\begin{array}{l} \sqrt{c^{2}+4+d^{2}}+2=r \\ \sqrt{(c-\sqrt{21})^{2}+d^{2}}+3=r \\ \sqrt{(c-\frac{\sqrt{21}}{7})^{2}+(d-\frac{12\sqrt{7}}{7})^{2}}+3=r \\ \end{array} \right.\] Solving $M(\frac{4\sqrt{21}}{7},0,\frac{6\sqrt{7}}{7})$ and $r=6$. but you do not give the final answer to this problem...

Midpoint $M$ of the sphere, externally tangent to all those four spheres. Radius $r=6$.

vanstraelen wrote: Midpoints of the spheres $M_{1}(0,-2,0),M_{2}(0,2,0),M_{3}(\sqrt{21},0,0),M_{4}(\frac{\sqrt{21}}{7},0,\frac{12\sqrt{7}}{7})$, see two pictures. $M_{1}$ green, $M_{2}$ red, $M_{3}$ blue, $M_{4}$ purple. So $M_{1}M_{2}=4,M_{1}M_{3}=4,M_{1}M_{4}=5,M_{2}M_{3}=5,M_{2}M_{4}=5,M_{3}M_{4}=6$. Tetrahedron $(M_{4},M_{1}M_{2}M_{3})$. Midpoint $M(c,0,d)$ and radius $r$, satisfying: \[\left\{\begin{array}{l} MM_{2}+2=r \\ MM_{3}+3=r \\ MM_{4}+3=r \\ \end{array} \right.\] \[\left\{\begin{array}{l} \sqrt{c^{2}+4+d^{2}}+2=r \\ \sqrt{(c-\sqrt{21})^{2}+d^{2}}+3=r \\ \sqrt{(c-\frac{\sqrt{21}}{7})^{2}+(d-\frac{12\sqrt{7}}{7})^{2}}+3=r \\ \end{array} \right.\] Solving $M(\frac{4\sqrt{21}}{7},0,\frac{6\sqrt{7}}{7})$ and $r=6$. There are some errors in your solution: $M_{1}M_{3}$ should be $5$; and since the fifth sphere is externally (not internally) tangent to the other spheres, hence $MM_{2}-2=r, MM_{3}-3=r , MM_{4}-3=r,$ ... Consequently, $6$ is not the right answer.

Recalculating $r_{1}=\frac{6}{11}$.

vanstraelen wrote: Recalculating $r_{1}=\frac{6}{11}$. Finally you get it right.