The equality in the statement involves many squares. The first thing to do is to study it modulo a number $n$ for which there are few quadratic residues. We therefore study the case $n=8$, as the squares modulo 8 are 0, 1, and 4.
In particular, if $p$ is odd, the equation becomes $a^{2}+b^{2}+c^{2} \equiv-1(\bmod 8)$, and this equation has no solutions. We deduce that $p=2$, and it is now a matter of finding the integers $a, b$, and $c$ such that $298=9 a^{2}+17 b^{2}+17 c^{2}$.
Without loss of generality, we assume that $a, b$, and $c$ are positive or zero. Since $b$ and $c$ play symmetric roles, we assume even that $b \leqslant c$. We then observe that $9 \times 6^{2}=324 > 298 \geqslant 9 a^{2}$, so that $0 \leqslant a \leqslant 5$. Since $9 a^{2}+17 b^{2}+17 c^{2} \equiv 298 \equiv 9(\bmod 17)$, we deduce in addition that 17 divides $9(a-1)(a+1)$. This means that $a=1$, and our equation becomes $17=b^{2}+c^{2}$, which means that $b=1$ and $c=4$.
In conclusion, the only possible solutions are

$$(a, b, c, p)=( \pm 1, \pm 1, \pm 4,2) \text { and }( \pm 1, \pm 4, \pm 1,2)$$

Conversely, and by virtue of the equality $73 \times 2^{2}+6=298=9 \times 1^{2}+17 \times 1^{2}+17 \times 4^{2}$, these quadruplets indeed work.

Comment from the graders: Many students approached the problem but few managed to make real progress. The exercise consisted of two parts: looking modulo 8 to show that $p=2$, then finding the solutions. It is a pity that many students attempted simplifications without ultimately looking for the case $p=2$, which was quite easy to handle and could have earned points. This case is also more relevant than cases like $a=b$ or $a=b=c$, since there is little chance of achieving such an equality.
Regarding the modular studies of the first equation, looking modulo 8 is quite natural in the presence of squares. When there are multiple squares present, one should study modulo 4, 3, and then 8 first, as these are three moduli very well suited for these equations. To succeed, one needs to look at what the squares modulo 8 are, which are 0, 1, and 4, and deduce that $p$ cannot be odd.
Let's also recall that it is pointless to look at the equation modulo a number $n$ that has two different prime factors, such as $n=6$ or $n=10$. For $n=6$, the equation provides as much information as the equation taken modulo 2 and then modulo 3, and the calculations are simpler in this case.