Let's determine $p$ such that the following equation

$$
3 x^{2}-5(p-1) x+p^{2}+2=0
$$

has roots that satisfy the following relationship:

$$
x_{1}+4 x_{2}=14
$$
According to our task,

$$
x_{1}+4 x_{2}=14
$$

$$
\frac{5(p-1)}{3}=x_{1}+x_{2}
$$

$$
p^{2}+2=3 x_{1} x_{2}
$$

If we substitute the value of $x_{1}=14-4 x_{2}$ from (1) into (2) and (3), we get:

$$
\frac{5}{3}(p-1)=14-3 x_{2}
$$

and

$$
p^{2}+2=3 x_{2}\left(14-4 x_{2}\right) \text {. }
$$

Eliminating $x_{2}$ from these two equations, we obtain:

$$
127 p^{2}-1250 p+2968=0
$$

from which

$$
p_{1}=\frac{742}{127} \quad \text { and } \quad p_{2}=4
$$