jmc

We take cases on the value of $x.$ If $x\le 1,$ then we have $(1-x)=(2-x)+(3-x),$ so $x=4.$ But this does not satisfy $x<1,$ so it is not a valid solution.

If $1<x\le 2,$ then we have $x-1=(2-x)+(3-x),$ so $x=2,$ which is a valid solution.

If $2<x\le 3,$ then we have $x-1=(x-2)+(3-x),$ so $x=2$ again.

If $3<x,$ then we have $(x-1)=(x-2)+(x-3),$ which gives $x=4.$ This time, $x=4$ is a valid solution because it satisfies $3<x.$

Hence there are $\boxed{2}$ values of $x,$ namely $x=2$ and $x=4.$