To intuit what is happening, interpret the Bernoulli$(1/2)$ results as coin flips, and so an XOR result is the event of getting exactly one head on a flip of two coins.
$$\mathsf P(X) = \mathsf P(B_1\oplus B_2) = \tfrac 1 2$$
And likewise for the rest.
$X$ and $Y$ have one coin in common, so the intersection of these events is that of either getting a head on only both uncommon coins (and a tail on the common one) xor of getting a head only on the common coin.
$$(B_1\oplus B_2)\cap (B_2\oplus B_3) = (B_1\cap B_3\cap B_2^\complement)\oplus (B_2\cap B_1^\complement\cap B_3^\complement)$$
$$\therefore \mathsf P(X\cap Y) = ?$$
Thus we have pairwise independence on $X$ and $Y$ if $\mathsf P(X\cap Y) = \mathsf P(X)~\mathsf P(Y)$. Does it?
By symmetry the same holds for $(X, Z)$ and $(Y, Z)$.
Now, what about mutually independence?