Sum of Pandigital Triplet of 3-Digit Primes

Theorem
The smallest integer which is the sum of a set of $3$ three-digit primes using all $9$ digits from $1$ to $9$ once each is $999$:
 * $149 + 263 + 587 = 999$

Proof
All three-digit primes end in $1, 3, 7, 9$.

Suppose $1$ is used as the units digit of a prime.

Since the digit $1$ cannot be used again, the sum of the primes is at least:
 * $221 + 333 + 447 = 1001$

so $1$ cannot be used as a units digit.

The units digits of the primes are $3, 7, 9$.

To minimise the sum, the hundreds digits must be $1, 2, 4$.

This leaves $5, 6, 8$ be the tens digits.

The primes satisfying these conditions are:
 * $157, 163, 167$
 * $257, 263, 269, 283$
 * $457, 463, 467, 487$

Only $269$ contain $9$, so we must choose it.

Only $157$ does not contain $6$, so must choose it next.

But then all primes beginning with $4$ have some digit coinciding with the above primes.

Hence the next minimal sum of the primes (if they exist) is:
 * $10^2 \paren {1 + 2 + 5} + 10 \paren {4 + 6 + 8} + \paren {3 + 7 + 9} = 999$

and we have shown that these primes do exist.