A Triple Prime

Since $0< X, Y, Z<10$ and all are prime we have $X,Y,Z\in\{2,3,5,7\}$

If any two are $2$ and $5$ the last digit of $\overline{XYZ}$ is $0$ which is not possible value.

We have $(X,Y,Z)$ is a permutation of $2,3,7$ or $3,5,7$ , by divisibility rules for $2$ and $5$ each of these cases has $Z=2$ and $Z=5$ respectively.

All permutations are already divisible by $3$ becasue $2+3+7=12$ and $3+5+7=15$ are both mutliples of $3$

Divisibility rule for $7$ says subtract $2$ times the last digit from the others

For $Z=2$ we have $37-4=33$ and $73-4=69$ , neither are divisble by $7$

For $Z=5$ we have $37-10=27$ and $73-10=63$ and $7\mid63$

So we have $\overline{XY}=73$ and subsequently $\overline{XYZ}=\boxed{735}$

Short Python Solution. Directly yield output : 735

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def prime(i):
    if i<2:
        return False
    for j in range(2,i):
        if i%j==0:
            return False
    return True
for i in range (100,1001):
    j=i
    a=j%10
    j//=10
    b=j%10
    j//=10
    c=j
    if prime(a) and prime(b) and prime(c) and i%a==0 and i%b==0 and i%c==0 and a!=b and b!=c:
        print(i)

Output:
735

There are 4 primes to use for XYZ

2357

The number has to be divisible by 3 of them, which means it has them as factors

210=2 3 5*7

210/2 = 105 as base? maybe

210/3 = 70 as base? no, zero rules it out

210/5 = 42 as base? maybe

210/7 = 30 as base? no zero rules it out.

Take the numbers 2 & 5...

multiples of 5 end in either 5 or 0, only 5 is prime

multiples of 2 can be 2 4 6 8 and 0, only 2 is prime

This means that it has to end with either 2 or 5, but with 210*N added on to it.

42? no

252? no

462? no

672? no

882? no

1092? no

105? no

315? no

525? no

735? yes! <---

945? no

There might be a neat way of doing this but I am going to go for a longer, more accessible method called casework.It is important to note that the primes available are 2,3,5 and 7. Now lets start by looking at the last digit Z. If Z=2, then we have 6 possibilities: (X,Y) = (3,5) , (3,7) , (5,7) + reverse order. However, if X or Y equals 3 then the digit sum must be divisible by 3 (divisibility by 3 rule) so (3,5) & (5,3) aren't valid. Also, X or Y can't equal 5 otherwise Z= 5, which contradicts Z=2. This excludes all possibilities for Z=2.

For Z=5, we have (X,Y) = (2,3) , (2,7) , (3,7) +reverse. Now using the divisibility by 3 rule (3,7) & (7,3) are the only possibilities left. Testing both gives XYZ = $\boxed{735}$ = 3 $\times\ 5 \times\ 7^{2}$ . Now we just need to verify that there are no more solutions.

For Z=3: we use the divisibility by 3 rule again to exclude (2,5) & (5,2). Now if 5 is a digit then 5| XYZ but that implies that Z=5 so (5,7) &(7,5) are also excluded. (2,7) and (7,2) are excluded as well as XYZ is odd. So no solutions for Z = 3.

For Z=7: (2,5), (3,5) + reverse are excluded as 5 doesn't divide XYZ unless Z=5. Similarly (2,3) +reverse is excluded as 2 can't divide an odd number.