This problem was proposed by Moldova

If $p>2$ $\pmod{8}$ gives us $7=a^2+b^2+c^2$, which is impossible since $a^2\in \{0,1,4\}\pmod{8}$. If $p=2$ we have $298=9a^2+17b^2+17c^2$. Note that $b^2+c^2\leq \frac{298-9\cdot 1^2}{17}=17$, which in particular implies $1\leq b,c\leq 4$. The equality case gives two solutions $(1,1,4,2)$ and $(1,4,1,2)$. Taking $\pmod 9$ we see $b^2+c^2\equiv 8$, and since $b^2\in \{0,1,4,7\}\pmod 9$ there are o my two possible configurations: 1)$b=4,c=1$ which we've already seen. 2) $b=c=2$ which doesn't work since $a^2=\frac{298-17\cdot 8}{9}=18$. Therefore the only two possible solutions are $$\{(1,1,4,2),(1,4,1,2)\}$$

For $p=2$, @above have pointed it out! Now, we will just prove that for $p\geq3$, there is no solution. Taking $\pmod 4$, we have that $p^2+2\equiv a^2+b^2+c^2\pmod4$, however $p^2\equiv 1\pmod4$ so, $a^2+b^2+c^2\equiv 3\pmod4$ and, $a^2,b^2,c^2\equiv1 \pmod4$. Since $a^2\equiv 1 \pmod 4$ then, $a^2\equiv 1 \pmod8$, and $a^2+b^2+c^2\equiv 3 \pmod 8$. Now, taking $\pmod 8$ we have, \[p^2+6\equiv a^2+b^2+c^2\equiv 3 \pmod 8\]so, $p^2\equiv 5 \pmod 8$, a contradiction! So, only $p=2$ satisfy the problem!

If $p$ is an odd prime, then $p^2\equiv 1 \pmod{8}$, and taking both sides of the equation modulo $8$ gives $p^2+6\equiv7\equiv a^2+b^2+c^2$, which is clearly impossible, as all squares are either $0$, $1$, or $4$, modulo $8$. Thus, $p=2$. Then, the equation becomes $298 = 17^2 + 9 = 9a^2 + 17b^2 + 17c^2$. Note that $a<6$ for size reasons. The equation can be rearranged as $17(17-b^2-c^2) = 9(a^2-1)$, which implies $a^2\equiv 1 \pmod{17}$, or $a\equiv 1,-1\pmod{17}$. Thus, $a=1$, and we have $17 = b^2+c^2$. Inspection gives that $(b,c)$ is a permutation of $(1,4)$, so the two solutions to the original equation are $(a,b,c,p) = \boxed{(1,1,4,2), (1,4,1,2)}$.