The sampling without replacement

You can do it this way but it is a bit clumsy:

For $ \left | x2-x1 \right |=0$

Probability[x == 2 || y == 2 || z == 2, {x, y, z} \[Distributed]
MultivariateHypergeometricDistribution[2, {3, 3, 4}]]

For $ \left | x2-x1 \right |=1$

Probability[(x == 1 && y == 1) || (y == 1 && z == 1), {x, y, z}
\[Distributed]MultivariateHypergeometricDistribution[2, {3, 3, 4}]]

For $ \left | x2-x1 \right |=2$

Probability[(x == 1 && z == 1), {x, y, z}
\[Distributed]MultivariateHypergeometricDistribution[2, {3, 3, 4}]]

I am a bit late but another way to use MultivariateHypergeometricDistribution:

md = MultivariateHypergeometricDistribution[2, {3, 3, 4}];
f[{___, 2, ___}] := 0
f[{1, 0, 1}] := 2
f[{___, 1, 1, ___}] := 1
td = TransformedDistribution[
   f[{x, y, z}], {x, y, z} \[Distributed] md];
res = Probability[x == #, x \[Distributed] td] & /@ Range[0, 2]
Histogram[RandomVariate[td, 10000], Automatic, "PDF", 
 Epilog -> {Red, PointSize[0.02], 
   Point[MapIndexed[{#2[[1]] - 1/2, #1} &, res]]}]

Amplifying on answer by @bobbym

p[x_] = Piecewise[{Probability[#[[1]], {x, y, z} \[Distributed] 
       MultivariateHypergeometricDistribution[2, {3, 3, 4}]], 
     x == #[[2]]} & /@
   {{x == 2 || y == 2 || z == 2, 0},
    {(x == 1 && y == 1) || (y == 1 && z == 1), 1},
    {x == 1 && z == 1, 2}}]

The distribution for |x1-x2| is then

dist = ProbabilityDistribution[p[x], {x, 0, 2, 1}];

This distribution can be used like any other distribution

PDF[dist, x] // Simplify

CDF[dist, x]

Mean[dist]

(*  1  *)

Variance[dist]

(*  8/15  *)

SeedRandom[1]

RandomVariate[dist, 10]

(*  {2, 0, 2, 0, 0, 0, 1, 0, 1, 1}  *)