Assuming that "(X1,X2 ⊥ X3)" (which makes no sense to me)
is a typo for (X1,X2 | X3), and translating the rest of
your jargon from an unspecified source as asking
"Find an example of three random variables $X,Y,Z$ that are pairwise independent but for which $X$ and $Y$ are NOT conditionally independent given $Z$",
see this answer of mine
for an example of three Bernoulli random variables that have this property as well as an example of three normal random variables that
that have this property.
A different example can be found in this answer on math.SE where $(X,Y,Z)$ is
uniformly distributed on four of the eight identical subcubes into which we can partition the cube $[0,1]^3$. The subcubes are
$[0,\frac 12]^3$ and $[\frac 12,1]^2\times[0,\frac 12]$ which sit on the $z=0$ plane.
$[0,\frac 12]\times [\frac 12,1]^2$ and
$[\frac 12,1]\times[0,\frac 12]\times[\frac 12,1]$ which sit on the $z=\frac 12$ plane.
In the answer on math.SE, it is shown that $X, Y, Z$ are pairwise
independent $\mathcal U([0,1]$ random variables, that is,
$$f_{X,Y}(x,y) = f_{X}(x)f_{Y}(y) = \begin{cases}1
& 0 \leq x, y \leq 1,\\0, & \text{otherwise}\end{cases}.$$ and similarly for the other two pairs. In particular, note that the pairwise
independence of $X$ and $Z$ means that $f_{X\mid Z}(x\mid z) = f_X(x)$ and similarly, we have that $f_{Y\mid Z}(y\mid z) = f_Y(y)$.
But, given that $Z = z \in [0,\frac 12]$, $(X,Y)$ is uniformly distributed on the set
$$\left(\left[0,\frac 12\right]\times\left[0,\frac 12\right]\right)
\bigcup\left(\left[\frac 12,1\right]\times\left[\frac 12,1\right]\right).$$
That is,
$$f_{X,Y\mid Z}(x,y\mid z)
= \begin{cases}2,& 0 \leq x,y \leq \frac 12,\\
2,& \frac 12 \leq x,y \leq 1,\\
0, &\text{otherwise},\end{cases}
\quad \neq f_{X\mid Z}(x\mid z)f_{Y\mid Z}(y\mid z)$$
showing that $X$ and $Y$ are NOT conditionally independent for
any given value of $Z$ in $[0,\frac 12)$, and a similar calculation
holds when the value of $Z$ is in $(\frac 12, 1]$.
