Squared Primes (Olympiad problem)

Look mod $3$ : get $a^2+b^2+c^2 \equiv 1$ mod $3$ . Since squares are $0$ or $1$ mod $3$ , this implies that two of the three primes are congruent to $0$ mod $3$ --that is, they must equal $3$ .

So we get $18 + 16c^2 = 9k^2+1$ or (up to interchanging $a$ and $b$ ) $153+b^2=9k^2+1$ .

The first equation gives $(3k-4c)(3k+4c) = 17$ , so $k = 3, c = 2$ is clearly the only possibility, so we've got the solution $(3,3,2,3)$ .

The second equation gives $(3k-b)(3k+b) = 152$ . The only two possibilities are that the factors are $4 \cdot 38$ and $2 \cdot 76$ because they have to have the same parity. This leads to $(3,7,3,17)$ and $(3,13,3,37)$ . Plus we can switch $a$ and $b$ to get $(7,3,3,17)$ and $(13,3,3,37)$ .

The sum of all the coordinates is $\fbox{183}$ .