Use that $24^2-210=(a+b+c)^2-a^2-b^2-c^2 =2(ab+bc+ca)$, thus $ab+bc+ca=183$. Now by Vieta the numbers are the three roots of $x^3-24x^2+183x-440$, thus $\{a,b,c\}=\{5,8,11\}$.

Since $ a^{2}+b^{2}+c^{2}\equiv2\pmod4$, one of $ a,b,c$ is even (wlog $ a$) and the other two are odd. Or, since $ 440 = 8\cdot55$, then $ \frac{a}{8}$ is an odd integer. Since $ 24^{2}> 210$, then $ |a| = 8$. If $ a =-8$, then $ b^{2}+c^{2}\geq\frac{(b+c)^{2}}{2}= 512$, absurd, so $ a = 8$, and $ b+c = 16$, $ |b-c| =\sqrt{b^{2}+c^{2}-2bc}= 6$, from where wlog $ b = 11$ and $ c = 8$. Because of symmetry between variables, all solutions are then a permutation of $ \left(5,8,11\right)$.

This solution solve integer solution but as Zetax we solve real solution . I think it is more interesting.

I totally agree with you, TTsphn, I just wanted to show an alternate solution without going through a cubic equation, and using properties of integers, that is all. I do not consider my solution an improvement, just an alternative that may (or may not) be useful in other cases where the equations are not so "symmetric", such as the following: $ 2a+b+4c=59,$ $ a^{2}+5b^{2}+c^{2}=310,$ $ abc=440.$ I cannot see a cubic equation coming up on this one, but the principle of my solution is still valid, as shown below: Again there is one of $ a$,$ b$,$ c$ that must be even, and two that must be odd, since $ a^{2}+b^{2}+c^{2}=310-4b^{2}\equiv2(mod4)$, so one of $ a$,$ b$,$ c$ is an odd multiple of 8. But it cannot be $ b$, since $ 5\cdot8^{2}=320>310$, and it must be either $ |a|=8$ or $ |c|=8$, since $ 24^{2}=576>310$. If $ c=-8$, then $ 2a+b=91$, $ ab=-55$, or $ |2a-b|=\sqrt{91^{2}+440}$, not an integer, while if $ c=8$, then $ 2a+b=27$, and $ |2a-b|=\sqrt{27^{2}-440}=17$, yielding either $ b=5$ and $ a=11$, or $ a=\frac{5}{2}$, i.e., not an integer solution. If $ a=-8$, then $ b+4c=75$, $ bc=-55$, $ |b-4c|=\sqrt{75^{2}+880}$, not an integer, while if $ a=8$, then $ b+4c=43$, $ |b-4c|=\sqrt{43^{2}-880}$, again not an integer... However, I still completely agree that, in the symmetric case, Zetax's solution is MUCH more elegant than mine... [/hide]

Another Solution

Knowing that $a,b,c$ are integers, and $abc=440$, it follows that one of them is divisible by $11$. Suppose wlog that $11|a$; it must be taken into account the fact that if $a^2+b^2+c^2=210<22^2$, then $|a|=11$, yielding two possibilities for b and c: $b+c=35$ $\rightarrow$ $bc=-40$ $b+c=13$ $\rightarrow$ $bc=40$ But $b^2+c^2=(b+c)^2-2bc=89$ cannot hold if $b+c=35$, so we rule out that possibility. Substituting $b=\frac{40}{c}$, we get that $c^2-13c+40=0$, with $8$ and $5$ as roots, so the triples $(a,b,c)$ are $(5,8,11)$ and all its permutations.

abc = 440 is a multiple of 11. WLOG let a be a multiple of 11. If a》22, then a^2 + b^2 + c^2 =210 is not possible. So, a = 11. Hence, bc = 40 & b+ c = 13. So, b = 5 & c = 8 is a possibility. All other possibilities are permutations of 5, 8 & 11.

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=576$ We can easily get that $ab+bc+ca=183$ and $ab=\frac{440}{c}$ and $a+b=24-c$, so $\frac{440}{c}+c(24-c)=183$. Now we will solve this simple equation $440+c^2(24-c)=183c$ $440+24c^2-c^3=183c$, and after factirazation you will get $(c-5)(c-8)(c-11)=0$, so $c$ can be $5,8,11$ First case $c=5$: $a+b=19$, $ab=88$ $\implies$ $a=\frac{88}{b}$ Now, solve equation $\frac{88}{b}+b=19$ to get $b=8$ or $b=11$, so $a=11$ or $a=8$. We have two pairs here $\{a,b,c\}=\{11,8,5\},\{8,11,5\}$ Second case $c=11$: $a+b=13$, $ab=40$ $\implies$ $a=\frac{40}{b}$ Now, solve again this equation $\frac{40}{b}+b=13$ to get $b=5$ or $b=8$, so $a=8$ or $a=5$. We have two pairs here $\{a,b,c\}=\{8,5,11\},\{5,8,11\}$ Third and last case $c=8$: $a+b=16$, $ab=55$ $\implies$ $a=\frac{55}{b}$ Now, solve equation $\frac{55}{b}+b=16$ to get $b=5$ or $b=11$, so $a=11$ or $a=5$. We have two pairs here $\{a,b,c\}=\{11,5,8\},\{5,11,8\}$ Finally, all solutions are $\{a,b,c\}=\{11,8,5\},\{8,11,5\},\{8,5,11\},\{5,8,11\},\{11,5,8\},\{5,11,8\}$

At first, note that $(a+b+c)^2-2(ab+bc+ac)=a^2+b^2+c^2$, thus, by replacing values we get $24^2-2(ab+bc+ac)=210$. By solving this we get $ab+bc+ac=183$, so, for the polynomial $f(x)=(x-a)(x-b)(x-c)$ we have $f(x)=x^3-(a+b+c)x^2+(ab+bc+ac)x-abc=x^4-24x^3+183x-440$, hence, by finding the roots of this polynomial we'll get the solutions of the system of equations. To see find them, we just need to look at the number $440=2^3\cdot 5\cdot 11$, and we get $\{a, b, c\} = \{5, 8, 11\}$. .