2D-3D

$\textrm{Denote the points at which the circle touches the }x,\textrm{ }y\textrm{ and }z\textrm{ axes as }$

$(a,0,0),\textrm{ }(0,b,0)\textrm{ and }(0,0,c)\textrm{ respectively.}$

$\textrm{If the circle has radius }r\textrm{, then:}$

$\sqrt{5+(c-3)^2}=\sqrt{10+(b-2)^2}$

$\sqrt{10+(b-2)^2}=\sqrt{13+(a-1)^2}$

$\sqrt{13+(a-1)^2}=r\textrm{.}$

$\textrm{Furthermore, if we denote the side lengths of the triangle formed}$

$\textrm{by joining }(a,0,0)\textrm{ to }(0,b,0)\textrm{, }(0,b,0)\textrm{ to }(0,0,c)\textrm{ and }(0,0,c)\textrm{ to }$

$(a,0,0)\textrm{ as }A\textrm{, }B\textrm{ and }C\textrm{ respectively, we know that:}$

$A=\sqrt{a^2+b^2}$

$B=\sqrt{b^2+c^2}$

$C=\sqrt{c^2+a^2}\textrm{.}$

$\textrm{Finally, the radius of the circumcircle of the triangle with side}$

$\textrm{lengths }A\textrm{, }B\textrm{ and }C\textrm{ is given by the formula:}$

$r=\dfrac{ABC}{\sqrt{(A+B+C)(A+B-C)(A-B+C)(B-A+C)}}\textrm{.}$

$\textrm{Substituting }A\textrm{, }B\textrm{ and }C\textrm{ in terms of }a\textrm{, }b\textrm{ and }c\textrm{ results in the formation}$

$\textrm{of a fourth equation relating }a,\textrm{ }b,\textrm{ }c\textrm{ and }r\textrm{.}$

$\textrm{So we have a system of four equations and four variables which}$

$\textrm{can be solved to yield }r\approx 4.844150887\textrm{.}$

$\textrm{So, rounded to 3 decimal places, }r\approx\boxed{4.844}\textrm{.}$

This is really about solving the system of four equations

$\dfrac { 1 }{ a } +\dfrac { 2 }{ b } +\dfrac { 3 }{ c } =1$
${ \left( a-1 \right) }^{ 2 }+{ \left( 0-2 \right) }^{ 2 }+{ \left( 0-3 \right) }^{ 2 }={ r }^{ 2 }$
${ \left( 0-1 \right) }^{ 2 }+{ \left( b-2 \right) }^{ 2 }+{ \left( 0-3 \right) }^{ 2 }={ r }^{ 2 }$
${ \left( 0-1 \right) }^{ 2 }+{ \left( 0-2 \right) }^{ 2 }+{ \left( c-3 \right) }^{ 2 }={ r }^{ 2 }$

and there doesn't seem to be any nice way of solving this. One way would to find that

$a=1+\sqrt { { c }^{ 2 }-6c+1 }$
$b=2+\sqrt { { c }^{ 2 }-6c+4 }$

and then solve the following equation for $c$ numerically

$\dfrac { 1 }{ a } +\dfrac { 2 }{ b } +\dfrac { 3 }{ c } =1$

Once we have $c=7.29718487...$ , we compute $r=4.84415089...$