We can easily check that $p=+2$ is not a solution to the problem, since in this case $4 p+1=4 \cdot 2+1=9$ is not a prime number.

For $p=-2$, however, $2 p+1=-3, 4 p+1=-7, 6 p+1=-11$, which is suitable.

The question is whether there are any other solutions to the problem.

Notice that for any integer $p$, exactly one of $p, 2 p+1, 4 p+1$ is divisible by 3, while $6 p+1$ is never divisible by 3.

1. If $p$ is divisible by 3, then since it is a prime number, this can only happen if $p= \pm 3$.

If $p=-3$, then $2 p+1=-7, 4 p+1=-13, 6 p+1=-19$. If $p=3$, then $2 p+1=7, 4 p+1=13, 6 p+1=19$ and if $p=-3$, then $2 p+1=-5, 4 p+1=-11, 6 p+1=-17$.

In this case, all the obtained values are prime numbers, so $p= \pm 3$ is a solution.

2. If 3 is a divisor of $(2 p+1)$, then the possible prime values of $2 p+1$ are $2 p+1=3$, from which $p=1$, which is not a prime number, or $2 p+1=-3$, from which $p=-2$, and we have already seen that this is a solution.
3. Similarly, if $4 p+1=3$, then $p=\frac{1}{2}$ is not an integer, or $4 p+1=-3$ and $p=-1$, which is not a prime.

In summary: the sought values of $p$ are -2 and $\pm 3$, and there are no other solutions.
For what $p$ (not necessarily positive) primes will the expressions $2 p+1, 4 p+1, 6 p+1$ also be prime?