2. Determine all triples of positive integers $(a, b, c)$ such that each of $a b-c$, $b c-a$, and $c a-b$ is a power of 2 (a power of 2 is an integer of the form $2^{n}(n \in \mathbf{N})$).
2. If $a=1$, then $b-c$ and $c-b$ are both powers of 2, which is impossible.
Thus, $a \geqslant 2$.
Similarly, $b \geqslant 2, c \geqslant 2$.
We will discuss two cases below.
【Case 1】 Two of $a, b, c$ are equal.
Without loss of generality, assume $a=b$.
Then, $a c-b=a(c-1)$ is a power of 2. This implies that both $a$ and $c-1$ are powers of 2.
Let $a=2^{s}, c=1+2^{t} (s \geqslant 1, t \geqslant 0)$.
Then, $a b-c=2^{2 s}-2^{t}-1$ is a power of 2.
If $t>0$, then $2^{2 s}-2^{t}-1$ is odd, and it can only be $2^{2 s}-2^{t}-1=1 \Rightarrow 2^{t} \equiv 2(\bmod 4)$
$\Rightarrow t=1 \Rightarrow s=1$.
In this case, $a=b=2, c=3$.
If $t=0$, then $2^{2 s}-2$ is a power of 2, and it can only be $s=1$.
In this case, $a=b=c=2$.
It is easy to verify that $(2,2,2)$ and $(2,2,3)$ both satisfy the conditions.
【Case 2】 $a, b, c$ are all different.
Without loss of generality, assume $2 \leqslant a < b < c$.
We will further discuss two cases based on the size of $a$.
(1) $a=2$.
First, we prove that $\gamma=0$.
Suppose $\gamma>0$, from equation (3) we know $c$ is even, and from equation (2) we know $b$ is also even. Thus, the left side of equation (1)
$b c-a \equiv 2(\bmod 4)$,

but $2^{\alpha} \equiv 0(\bmod 4)$, which is a contradiction.
Therefore, $\gamma=0$.
So equation (3) becomes $c=2 b-1$.
From equation (2) we get $3 b-2=2^{\beta}$, and modulo 3 shows that $\beta$ is even.
If $\beta=2$, then $b=2=a$, which is not valid.
If $\beta=4$, then $b=6, c=11$.
It is easy to verify that $(2,6,11)$ satisfies the conditions.
If $\beta \geqslant 6$, then $b=\frac{1}{3}\left(2^{\beta}+2\right)$.
Substituting into equation (1) we get
$9 \times 2^{\alpha}=9(b c-a)=9 b(2 b-1)-18$
$=(3 b-2)(6 b+1)-16$
$=2^{\beta}\left(2^{\beta+1}+5\right)-16$.
Since $\alpha>\beta \geqslant 6$, we have $2^{7} \mid 19 \times 2^{\alpha}$. But equation (4) is only divisible by $2^{4}$, not by $2^{5}$, which is a contradiction.
(2) $a \geqslant 3$.
Adding and subtracting equations (1) and (2) respectively, we get
$(a+b)(c-1)=2^{\alpha}+2^{\beta}$,
$(b-a)(c+1)=2^{\alpha}-2^{\beta}$.
Since one of $c-1$ and $c+1$ is not divisible by 4, we have $2^{\beta-1} \mid(a+b)$ or $2^{\beta-1} \mid(b-a)$.
From $2^{\beta}=a c-b \geqslant 3 c-b>2 c$, we know
$b<3 b$,

which is a contradiction.
Therefore, $a=3, c-2=b$.
Thus, $b=2^{\beta-1}-3, c=2^{\beta-1}-1$.
Substituting into equation (3) we get
$2^{\gamma}=a b-c=3\left(2^{\beta-1}-3\right)-\left(2^{\beta-1}-1\right)$
$=2^{\beta}-8$.
Thus, $\beta=4, b=5, c=7$.
It is easy to verify that $(3,5,7)$ also satisfies the conditions.
In summary, there are 16 sets of positive integer triples that satisfy the conditions: the three permutations of $(2,2,2)$ and $(2,2,3)$, the six permutations of $(2,6,11)$, and the six permutations of $(3,5,7)$.