Find all positive integers $a, b, c$ and $p$, where $p$ is a prime number, such that

$$
73 p^{2}+6=9 a^{2}+17 b^{2}+17 c^{2}
$$
Since the equation is symmetric with respect to the numbers $b$ and $c$, we assume that $b \geq c$.

If $p \geq 3$, then $p$ is an odd number. We consider the equation modulo 8. Since,

$$
73 p^{2}+6 \equiv 79 \equiv 7 \quad(\bmod 8)
$$

we get that

$$
a^{2}+b^{2}+c^{2} \equiv 7 \quad(\bmod 8)
$$

This cannot happen since for any integer $x$ we have that

$$
x^{2} \equiv 0,1,4 \quad(\bmod 8)
$$

Hence, $p$ must be an even prime number, which means that $p=2$. In this case, we obtain the equation

$$
9 a^{2}+17\left(b^{2}+c^{2}\right)=289
$$

It follows that $b^{2}+c^{2} \leq 17$. This is possible only for

$$
(b, c) \in\{(4,1),(3,2),(3,1),(2,2),(2,1),(1,1)\}
$$

It is easy to check that among these pairs only the $(4,1)$ gives an integer solution for $a$, namely $a=1$. Therefore, the given equation has only two solutions,

$$
(a, b, c, p) \in\{(1,1,4,2),(1,4,1,2)\}
$$