391. Find three consecutive prime numbers (not necessarily differing by 1 or 2 from each other), the sum of whose squares is also a prime number. List all solutions.
$\triangle$ Let's denote these prime numbers in ascending order as $p, q$, and $r$. We will consider three cases.

1) Suppose $p=2$.

Then the numbers $q$ and $r$ are odd, which means the sum $4+q^{2}+r^{2}$ is even. Therefore, the case $p=2$ is impossible.

2) Suppose $p=3$.

The next prime numbers after 3 are 5 and 7. Let's check:

$$
3^{2}+5^{2}+7^{2}=9+25+49=83.
$$

We get a prime number, which is what we needed.

3) Suppose $p>3$.

Then the numbers $q$ and $r$ are also greater than 3. Therefore, none of the numbers $p, q$, and $r$ are divisible by 3. However, the square of a natural number that is not divisible by 3, when divided by 3, can only leave a remainder of 1 (see the statement of problem 163 from § 6). We will have:

$$
p^{2}+q^{2}+r^{2}=(3 k+1)+(3 l+1)+(3 m+1)=3 k+3 l+3 m+3
$$

Since this sum is divisible by 3, the case $p>3$ is impossible.

Answer: $3,5,7$.