In this paper (Proposition 4) you can find statement :
If $p$ is a prime of the form : $p = 2q + 1$ for some odd prime $q$,
then $2$ is a primitive root modulo $p$ if and only if : $q \equiv 1, 5 \pmod 8$.
Is it true that :
If $p$ is a prime of the form : $p =3\cdot q \cdot (q+1)-1$ for some odd prime $q$,
then $3$ is a primitive root modulo $p$ if and only if : $q \equiv 1, 5 \pmod {12}$.
Legendre symbol for $3$ is :
$\left(\frac{3}{p} \right) = \begin{cases} ~~~1, & \text{if } :p \equiv 1,11 \pmod {12} \\ -1, & \text{if } : p \equiv 5,7 \pmod {12} \end{cases}$
It is obvious that necessary condition is : $q \equiv 1, 5 \pmod {12}$ since :
$3 \cdot(12t+7)^2+3(12t+7)-1 \equiv 11 \pmod {12}$
$3 \cdot(12t+11)^2+3(12t+11)-1 \equiv 11 \pmod {12}$
There is a theorem that states :
$a$ is a primitive root modulo $~p~$ iff $~~ord_p(a) = \phi(p)$ .
So , how to prove that : $~~ord_p(3) =3\cdot q \cdot (q+1)-2 $ ?
EDIT :
Conjecture is false . Smallest counterexample is : $35862917$