The circles having radii 1, 2, 3 touch each other externally, then the radius of the circle which cuts the three circles orthogonally is

Given that three circles having radii 1, 2, 3 touch each other externally. 

The circle having radical center at center and the length of tangent from the radical center to any circle as radius intersect the given circles orthogonally

This is incircle to the triangle whose vertices are centers of the given three triangles

For the given set of circles, the lengths of the sides formed by centers are $3,4,5$ as shown in the following figure

Hence, the triangle is right angled triangle

The length of inradius is 

   $\begin{array}{rcl}r& =& \frac{∆}{s}\\ & =& \frac{{\displaystyle \frac{1}{2}}·4·3}{\left({\displaystyle \frac{3+4+5}{2}}\right)}\\ & =& \frac{6}{6}\\ & =& 1\end{array}$

Therefore, the inradius is $\boxed{\mathbf{1}}$