Find the number

For $x \equiv 3 \mod {7}$ , then $x = 7n + 3$ , where $n$ is an integer and $6 \le n \le 64$ for $45 \le x \le 454$ .

Now, for $x$ to be a prime, $n$ cannot not be a multiple of $3$ , else $x$ is divisible by 3. $n$ cannot be odd, else $n$ is even and divisible by $2$ . Therefore $n$ must be even.

Starting with $6$ which is not acceptable as it is a multiple of $3$ , the next possible $n$ is $8$ and $7\times 8 +3 = 59$ which is a prime, but the sum of its digits is $5+9 = 14 > 12$ which is unacceptable. The next possible $n$ is $10$ and $7 \times 10 +3 =73$ also a prime and the sum of digits $7+3 = 10 < 12$ . Therefore the answer is $\boxed{73}$