We can do this systematically quite easily. Case 1: $ a^{2}+b^{2}+c^{2}= 1$. Since $ a \neq 0$, 100 is the only valid number. Case 2: $ a^{2}+b^{2}+c^{2}= 2$. a must be 1, so 110 and 101 are the only valid numbers Case 3: $ a^{2}+b^{2}+c^{2}= 13$. Clearly, a,b,c must be 3 or less, and we see that one of the numbers must be 3. The only integral solution to $ x^{2}+y^{2}= 4$ is (2,0). So, the valid numbers are 320, 302, 230, 203. Case 4: 26. Triplets (x,y,z) such that $ x^{2}+y^{2}+z^{2}= 26$ are what we want to find. If one is 5, then we only have (5,1,0). If one is 4, we only have (4,3,1). If none are 4, then at least one must be 3, and we find that there are no integral solutions to $ i^{2}+j^{2}= 17$ for i,j < 4. So, the valid numbers are 510, 501, 150, 105, and 431, 413, 314, 341, 143, 134. And we are done.