The equation we are given is: $a^{2}+b^{2}+c^{2}=7-ab-bc-ac$

Isolating the variables on LHS we attain, $a^{2}+b^{2}+c^{2}+ab+bc+ac=7$

We can multiply the equation by 2 to obtain, $2a^{2}+2b^{2}+2c^{2}+2ab+2bc+2ac=14$

And hence factorized nicely into, $(a+b)^2+(b+c)^2+(a+c)^2=14$

Note that 14 is simply a sum of 3 distinct squares: $1^{2}, 2^{2} , 3^{2}$ .

Since $0 \le a < b < c$ , $(b+c)^2$ is largest, followed by $(a+c)^2$ and smallest is $(a+b)^2$ .

Thus, $(b+c)^2 = 3^2$ then, $b+c =3 \tag{1}$

Since, $(a+c)^2 = 2^2$ then, $a+c =2 \tag{2}$

Since, $(a+b)^2 = 1^2$ then, $a+b = 1 \tag{3}$

Adding $(1)+(2)+(3)$ we obtain, $2a+2b+2c = 6$ and hence,

$\boxed{a + b + c = 3}$

End of solution.

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Extras-- the only solution to this question for $(a,b,c)$ is $(0,1,2)$ .

$(2) + (3)$ we obtain: $2a+b+c=3$

Substituting $(1)$ in the equation above we get, $2a+3=3$ which leads to $a=0$ , solving $(2) , (3)$ we see that $b=1$ and $c =2$ .

Note: the question only asks for integers $a,b,c$ .